Triangles: 5 marks Questions (CBSE Class 10)

59 different 5 marks questions on Triangles from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (70)2 marks (56)3 marks (21)4 marks (4)5 marks (59)

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2026 · Basic 430/4/1

AD and PS are respectively, the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , then prove that
(i) $△ A D C \sim △ P S R$
(ii) $\frac{A D}{P S} = \frac{B C}{QR}$

Diagram for CBSE 2026 Class 10 Maths question 33 (OR)

Show answer & solution

Answer: Proved.

$△ A B C \sim △ P QR$ gives $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R}$ , $∠ B = ∠ Q$ and $∠ C = ∠ R$ .
D and S are midpoints, so DC = $\frac{1}{2}$ BC and SR = $\frac{1}{2}$ QR; hence $\frac{D C}{S R} = \frac{B C}{QR} = \frac{A C}{P R}$ .
(i) In $△ A D C$ and $△ P S R$ : $\frac{A C}{P R} = \frac{D C}{S R}$ and $∠ C = ∠ R$ , so $△ A D C \sim △ P S R$ (SAS similarity).
(ii) Similarly BD = $\frac{1}{2}$ BC and QS = $\frac{1}{2}$ QR give $\frac{A B}{P Q} = \frac{B D}{QS}$ with $∠ B = ∠ Q$ , so $△ A B D \sim △ P QS$ (SAS).
Hence $\frac{A D}{P S} = \frac{A B}{P Q} = \frac{B C}{QR}$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q325 marksLong AnswerTriangles CBSE 2026 · Basic 430/5/1

In the given figure, $△ A B C$ is right angled triangle with $∠ A = 9 0^{\circ}$ . AD is perpendicular to BC.
Prove that :
(i) $△ D B A \sim △ D A C$
(ii) $D A^{2} = D B \times D C$
(iii) Find the area of $△ A B C$ when DB = 9 cm and DC = 16 cm.

Diagram for CBSE 2026 Class 10 Maths question 32

Show answer & solution

Answer: (i) Proved. (ii) Proved. (iii) 150 cm

^{2}

(i) In $△ D B A$ and $△ D A C$ : $∠ A D B = ∠ C D A = 9 0^{\circ}$ .
$∠ D B A = 9 0^{\circ} - ∠ D A B = ∠ D A C$ (since $∠ D A B + ∠ D A C = 9 0^{\circ}$ ).
So $△ D B A \sim △ D A C$ (AA).
(ii) Corresponding sides are proportional: $\frac{D B}{D A} = \frac{D A}{D C}$ , so $D A^{2} = D B \times D C$ .
(iii) $D A^{2} = 9 \times 16 = 144$ , so DA = 12 cm; BC = 9 + 16 = 25 cm.
Area of $△ A B C = \frac{1}{2} \times B C \times A D = \frac{1}{2} \times 25 \times 12 = 150$ cm $^{2}$

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2026 · Standard 30/1/1

In the given figure, CM and RN are respectively the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , then prove that :
(i) $△ A M C \sim △ P N R$
(ii) $△ C M B \sim △ R N Q$

Diagram for CBSE 2026 Class 10 Maths question 34 (OR)

Show answer & solution

Answer: Proved.

Since $△ A B C \sim △ P QR$ : $∠ A = ∠ P$ , $∠ B = ∠ Q$ and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ .
M and N are mid-points, so $A M = M B = \frac{1}{2} A B$ and $P N = N Q = \frac{1}{2} P Q$ ; hence $\frac{A M}{P N} = \frac{M B}{N Q} = \frac{A B}{P Q}$ .
(i) In $△ A M C$ and $△ P N R$ : $\frac{A M}{P N} = \frac{A B}{P Q} = \frac{C A}{R P}$ and $∠ A = ∠ P$ .
So $△ A M C \sim △ P N R$ (SAS similarity).
(ii) In $△ C M B$ and $△ R N Q$ : $\frac{M B}{N Q} = \frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ .
So $△ C M B \sim △ R N Q$ (SAS similarity).

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q345 marksLong AnswerTriangles CBSE 2026 · Standard 30/2/1

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Given: In $△ A B C$ , DE $∥$ BC meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw EN $⊥$ AB and DM $⊥$ AC.
$a r (A D E) = \frac{1}{2} A D \cdot E N$ and $a r (B D E) = \frac{1}{2} D B \cdot E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
Similarly $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ (using height DM).
Triangles BDE and DEC are on the same base DE and between the same parallels DE and BC, so $a r (B D E) = a r (D E C)$ .
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2026 · Standard 30/2/1

As shown in the given figure, a girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Show answer & solution

Answer: 1.6 m

Distance walked in 4 s: BD = 1.2 × 4 = 4.8 m. Let shadow DE = x m. CD = 90 cm = 0.9 m, AB = 3.6 m.
$△ A B E \sim △ C D E$ (AA: right angles at B and D, common $∠ E$ ).
$\frac{A B}{C D} = \frac{B E}{D E}$ , so $\frac{3.6}{0.9} = \frac{4.8 + x}{x}$
$4 x = 4.8 + x$ , so $3 x = 4.8$ and x = 1.6
The shadow is 1.6 m long.

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q335 marksLong AnswerTriangles CBSE 2026 · Standard 30/3/1

In $Δ$ ABC, AD is a median. X is a point on AD such that AX : XD = 2 : 3. BX is extended so that it intersects AC at Y. Prove that BX = 4 XY.

Diagram for CBSE 2026 Class 10 Maths question 33

Show answer & solution

Answer: Proved.

Through D draw DZ $∥$ BY meeting AC at Z
In $△ C B Y$ , D is the midpoint of BC and DZ $∥$ BY, so DZ $= \frac{1}{2}$ BY
In $△ A D Z$ , XY $∥$ DZ, so $△ A X Y \sim △ A D Z$ (AA)
$\frac{X Y}{D Z} = \frac{A X}{A D} = \frac{2}{5}$ , so XY $= \frac{2}{5}$ DZ $= \frac{2}{5} \times \frac{1}{2}$ BY $= \frac{1}{5}$ BY
BX = BY − XY $= 5$ XY − XY $= 4$ XY
Hence BX = 4 XY

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q335 marksLong AnswerTriangles CBSE 2026 · Standard 30/5/1

D is the mid-point of side BC of $△ A B C$ . CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that
(i) OBGC is a parallelogram.
(ii) $E F ∥ B C$
(iii) $△ A E F \sim △ A B C$

Show answer & solution

Answer: Proved.

(i) BD = DC (D is the mid-point of BC) and OD = DG (given). So the diagonals BC and OG of quadrilateral OBGC bisect each other. Hence OBGC is a parallelogram.
(ii) As OBGC is a parallelogram, $O C ∥ B G$ and $O B ∥ C G$ , i.e. $O E ∥ B G$ and $O F ∥ C G$ .
In $△ A B G$ , $O E ∥ B G$ , so by BPT $\frac{A E}{E B} = \frac{A O}{O G}$ .
In $△ A C G$ , $O F ∥ C G$ , so by BPT $\frac{A F}{F C} = \frac{A O}{O G}$ .
Hence $\frac{A E}{E B} = \frac{A F}{F C}$ , and by the converse of BPT in $△ A B C$ , $E F ∥ B C$ .
(iii) In $△ A E F$ and $△ A B C$ : $∠ A$ is common, and $∠ A E F = ∠ A B C$ (corresponding angles, $E F ∥ B C$ ).
Hence $△ A E F \sim △ A B C$ (AA similarity).

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2026 · Standard 30/5/1

Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

Show answer & solution

Answer: Proved.

(i) In $△ A D Q$ and $△ R C Q$ : DQ = CQ (Q is the mid-point of CD), $∠ A Q D = ∠ R QC$ (vertically opposite), $∠ A D Q = ∠ R C Q$ (alternate angles, $A D ∥ B R$ ).
So $△ A D Q ≅ △ R C Q$ (ASA), hence AQ = QR.
(ii) In $△ A P B$ and $△ QP D$ : $∠ P A B = ∠ P Q D$ and $∠ P B A = ∠ P D Q$ (alternate angles, $A B ∥ D C$ ).
So $△ A P B \sim △ QP D$ (AA), and $\frac{A P}{P Q} = \frac{A B}{D Q}$ .
$D Q = \frac{1}{2} C D = \frac{1}{2} A B$ , so $\frac{A P}{P Q} = 2$ , i.e. AP = 2PQ.
(iii) PR = PQ + QR = PQ + AQ (from (i)) = PQ + (AP + PQ) = AP + 2PQ = AP + AP = 2AP (using (ii)).

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q335 marksLong AnswerTriangles CBSE 2025 · Basic 430/1/1

State “Basic Proportionality Theorem” and use it to prove the following :
In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{A O}{B O} = \frac{C O}{D O}$ as shown in the given figure. Prove that ABCD is a trapezium.

Diagram for CBSE 2025 Class 10 Maths question 33

Show answer & solution

Answer: Proved.

BPT: if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, it divides those two sides in the same ratio.
Proof of BPT: in $△ A B C$ let DE $∥$ BC meet AB at D and AC at E. Join BE and CD and draw DM $⊥$ AC, EN $⊥$ AB.
ar(ADE) $= \frac{1}{2} A D \cdot E N$ and ar(BDE) $= \frac{1}{2} D B \cdot E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ ; similarly $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
Triangles BDE and DEC are on the same base DE between the same parallels, so ar(BDE) = ar(DEC). Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .
Now in ABCD, draw OE $∥$ AB meeting AD at E.
In $△ A B D$ , OE $∥$ AB, so by BPT $\frac{A E}{E D} = \frac{B O}{O D}$ .
Given $\frac{A O}{B O} = \frac{C O}{D O}$ , i.e. $\frac{A O}{C O} = \frac{B O}{D O}$ . So $\frac{A E}{E D} = \frac{A O}{O C}$ .
In $△ A D C$ , E and O divide AD and AC in the same ratio, so OE $∥$ DC (converse of BPT).
Thus AB $∥$ OE $∥$ DC, so AB $∥$ DC and ABCD is a trapezium.

Q325 marksLong AnswerTriangles CBSE 2025 · Basic 430/1/2

State “Basic Proportionality Theorem” and use it to prove the following :
A line through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Show answer & solution

Answer: Proved.

BPT: if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, it divides those two sides in the same ratio.
Proof of BPT: in $△ A B C$ let DE $∥$ BC meet AB at D and AC at E. Join BE and CD and draw EN $⊥$ AB, DM $⊥$ AC.
$\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} A D \cdot E N}{\frac{1}{2} D B \cdot E N} = \frac{A D}{D B}$ and $\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} A E \cdot D M}{\frac{1}{2} E C \cdot D M} = \frac{A E}{E C}$ .
ar(BDE) = ar(DEC) (same base DE, between the same parallels DE and BC), so $\frac{A D}{D B} = \frac{A E}{E C}$ .
Application: in $△ A B C$ , let D be the mid-point of AB and let the line through D parallel to BC meet AC at E.
By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ .
Since AD = DB, $\frac{A E}{E C} = 1$ , so AE = EC.
Hence the line bisects the third side AC.

Q345 marksLong AnswerTriangles CBSE 2025 · Basic 430/1/3

State the converse of “Basic Proportionality Theorem” and use it to prove the following :
Line segment joining mid-points of any two sides of a triangle is parallel to the third side.

Show answer & solution

Answer: Proved.

Converse of BPT: if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Proof of the converse: let line DE meet AB at D and AC at E with $\frac{A D}{D B} = \frac{A E}{E C}$ , and suppose DE is not parallel to BC.
Draw DE' $∥$ BC meeting AC at E'. By BPT, $\frac{A D}{D B} = \frac{A E ^{'}}{E ^{'} C}$ , so $\frac{A E}{E C} = \frac{A E ^{'}}{E ^{'} C}$ .
Adding 1 to both sides: $\frac{A C}{E C} = \frac{A C}{E ^{'} C}$ , so EC = E'C and E, E' coincide. Hence DE $∥$ BC.
Application: in $△ A B C$ , let D and E be the mid-points of AB and AC.
Then AD = DB and AE = EC, so $\frac{A D}{D B} = 1 = \frac{A E}{E C}$ .
By the converse of BPT, DE $∥$ BC, i.e. the segment joining the mid-points is parallel to the third side.

Q335 marksLong AnswerTriangles CBSE 2025 · Basic 430/2/1

State the SAS criteria of similarity of two triangles. In the given figure, it is given that $O A \cdot O C = O B \cdot O D$ . Use the SAS criteria to prove that AD $∥$ CB.

Show answer & solution

Answer: Proved.

SAS similarity criterion: if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.
Given $O A \cdot O C = O B \cdot O D$ , so $\frac{O A}{O B} = \frac{O D}{O C}$ .
In $△ A O D$ and $△ B O C$ : $∠ A O D = ∠ B O C$ (vertically opposite angles) and $\frac{O A}{O B} = \frac{O D}{O C}$ .
So $△ A O D \sim △ B O C$ (SAS similarity).
Hence $∠ O A D = ∠ O B C$ , i.e. $∠ D A B = ∠ C B A$ .
These are alternate angles made by the transversal AB with lines AD and CB, so AD $∥$ CB.

Q325 marksLong AnswerTriangles CBSE 2025 · Basic 430/2/2

State AA criterion of similarity of two triangles and use it to prove the following.
In the given figures of $△ A B C$ and $△ P QR$ , AD and PS are angle bisectors of $∠ B A C$ and $∠ R P Q$ respectively.
If $△ A B C \sim △ P QR$ , prove that $△ A C D \sim △ P R S$ .

Diagram for CBSE 2025 Class 10 Maths question 32

Show answer & solution

Answer: Proved.

AA criterion: if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Since $△ A B C \sim △ P QR$ : $∠ A = ∠ P$ and $∠ C = ∠ R$ .
AD and PS bisect $∠ A$ and $∠ P$ , so $∠ C A D = \frac{1}{2} ∠ A = \frac{1}{2} ∠ P = ∠ R P S$ .
In $△ A C D$ and $△ P R S$ : $∠ C A D = ∠ R P S$ and $∠ A C D = ∠ P R S$ .
Hence $△ A C D \sim △ P R S$ (AA similarity).

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2025 · Basic 430/3/1

In the given figure, CM and RN are respectively, the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , prove that :
(i) $△ A M C \sim △ P N R$
(ii) $∠ B C M = ∠ QR N$
(iii) $△ B M C \sim △ QN R$

Diagram for CBSE 2025 Class 10 Maths question 33 (OR)

Show answer & solution

Answer: Proved.

Since $△ A B C \sim △ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}$ , $∠ A = ∠ P$ , $∠ B = ∠ Q$ .
(i) M and N are mid-points of AB and PQ, so $\frac{A M}{P N} = \frac{\frac{1}{2} A B}{\frac{1}{2} P Q} = \frac{A B}{P Q} = \frac{C A}{R P}$ .
With $∠ A = ∠ P$ , $△ A M C \sim △ P N R$ (SAS similarity).
(iii) Similarly $\frac{B M}{QN} = \frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ , so $△ B M C \sim △ QN R$ (SAS similarity).
(ii) From (iii), corresponding angles are equal, so $∠ B C M = ∠ QR N$ .

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2025 · Basic 430/4/1

In a $△$ ABC, P and Q are points on AB and AC respectively such that PQ $∥$ BC. Prove that the median AD, drawn from A to BC, bisects PQ.

Show answer & solution

Answer: Proved.

Let the median AD meet PQ at E. Then BD = DC.
In $△ A P E$ and $△ A B D$ : $∠ A$ is common and $∠ A P E = ∠ A B D$ (corresponding angles, PQ $∥$ BC). So $△ A P E \sim △ A B D$ (AA) and $\frac{P E}{B D} = \frac{A E}{A D}$ .
Similarly $△ A QE \sim △ A C D$ , so $\frac{QE}{C D} = \frac{A E}{A D}$ .
Hence $\frac{P E}{B D} = \frac{QE}{C D}$ ; since BD = CD, PE = QE.
So AD bisects PQ.

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2025 · Basic 430/5/1

It is given that sides AB and AC and median AD of $△ A B C$ are respectively proportional to sides PQ and PR and median PM of another $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ .
Produce AD to E with DE = AD and PM to N with MN = PM; join CE and RN.
$△ A B D ≅ △ E C D$ (BD = DC, AD = DE, vertically opposite angles), so CE = AB and $∠ B A D = ∠ C E D$ .
Similarly RN = PQ and $∠ QP M = ∠ R N M$ .
Now $\frac{C E}{R N} = \frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ .
So $△ A E C \sim △ P N R$ (SSS), giving $∠ C A E = ∠ R P N$ and $∠ C E A = ∠ R N P$ .
Then $∠ B A D = ∠ C E A = ∠ R N P = ∠ QP M$ .
Adding: $∠ B A C = ∠ B A D + ∠ D A C = ∠ QP M + ∠ M P R = ∠ QP R$ .
With $\frac{A B}{P Q} = \frac{A C}{P R}$ , $△ A B C \sim △ P QR$ (SAS).

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q335 marksLong AnswerTriangles CBSE 2025 · Standard 30/1/1

The diagonal $B D$ of a parallelogram $A B C D$ intersects the line segment $A E$ at the point $F$ , where $E$ is any point on the side $B C$ . Prove that $D F \times E F = F B \times F A$ .

Show answer & solution

Answer: Proved.

In parallelogram $A B C D$ , $A D ∥ B C$ , so $A D ∥ B E$ .
In $△ A F D$ and $△ E F B$ :
$∠ A F D = ∠ E F B$ (vertically opposite angles)
$∠ F A D = ∠ F E B$ (alternate angles, $A D ∥ B E$ , $A E$ transversal)
So $△ A F D \sim △ E F B$ (AA similarity).
Hence $\frac{D F}{B F} = \frac{F A}{F E}$ , which gives $D F \times E F = F B \times F A$ .

Also asked in: 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q345 marksLong AnswerTriangles CBSE 2025 · Standard 30/2/1

Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that $\frac{A M}{M B} = \frac{A N}{N D}$ where $L M ∥ C B$ and $L N ∥ C D$ .

Diagram for CBSE 2025 Class 10 Maths question 34

Show answer & solution

Answer: Proved.

Theorem (BPT): In $△ P QR$ , let $S T ∥ QR$ with S on PQ and T on PR. To prove $\frac{P S}{S Q} = \frac{P T}{T R}$ .
Join QT and RS. Draw $T N ⊥ P Q$ and $S M ⊥ P R$ .
$ar (△ P S T) = \frac{1}{2} \times P S \times T N$ and $ar (△ QS T) = \frac{1}{2} \times S Q \times T N$ , so $\frac{ar ( △ P S T )}{ar ( △ QS T )} = \frac{P S}{S Q}$
Similarly $\frac{ar ( △ P S T )}{ar ( △ R T S )} = \frac{P T}{T R}$
$△ QS T$ and $△ R T S$ are on the same base ST and between the same parallels ST and QR, so $ar (△ QS T) = ar (△ R T S)$
Hence $\frac{P S}{S Q} = \frac{P T}{T R}$ .
Application: In $△ A B C$ , $L M ∥ C B$ , so by BPT $\frac{A M}{M B} = \frac{A L}{L C}$
In $△ A D C$ , $L N ∥ C D$ , so by BPT $\frac{A N}{N D} = \frac{A L}{L C}$
Hence $\frac{A M}{M B} = \frac{A N}{N D}$

Also asked in: 2025 Standard 30/2/2, 2025 Standard 30/2/3

Q325 marksLong AnswerTriangles CBSE 2025 · Standard 30/3/1

The perimeter of an isosceles triangle is 32 cm. If each equal side is $\frac{5}{6}$ th of the base, find the area of the triangle.

Show answer & solution

Answer: 48 cm

^{2}

Let the base be $b$ cm; each equal side $= \frac{5 b}{6}$ cm.
$b + 2 \times \frac{5 b}{6} = 32$ , so $\frac{8 b}{3} = 32$ , $b = 12$ cm; equal sides $= 10$ cm.
The altitude to the base bisects it: height $= 1 0^{2} - 6^{2} = 64 = 8$ cm.
Area $= \frac{1}{2} \times 12 \times 8 = 48$ cm $^{2}$ .

Q345 marksLong AnswerTriangles CBSE 2025 · Standard 30/3/1

In the given figure, PA, QB and RC are perpendicular to AC. If PA = $x$ units, QB = $y$ units and RC = $z$ units, prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .

Show answer & solution

Answer: Proved.

PA, QB, RC are all perpendicular to AC, so $P A ∥ QB ∥ R C$ .
In $△ C A P$ , $QB ∥ P A$ , so $△ C B Q \sim △ C A P$ (AA) and $\frac{QB}{P A} = \frac{B C}{A C}$ , i.e. $\frac{y}{x} = \frac{B C}{A C}$ ... (1)
In $△ A C R$ , $QB ∥ R C$ , so $△ A B Q \sim △ A C R$ (AA) and $\frac{QB}{R C} = \frac{A B}{A C}$ , i.e. $\frac{y}{z} = \frac{A B}{A C}$ ... (2)
Adding (1) and (2): $\frac{y}{x} + \frac{y}{z} = \frac{B C + A B}{A C} = \frac{A C}{A C} = 1$ .
Dividing by $y$ : $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2025 · Standard 30/3/1

Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ , where D and M are mid-points of BC and QR.
$B D = \frac{1}{2} B C$ and $QM = \frac{1}{2} QR$ , so $\frac{B C}{QR} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$ .
Hence $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ , so $△ A B D \sim △ P QM$ (SSS).
Therefore $∠ B = ∠ Q$ .
In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ .
So $△ A B C \sim △ P QR$ (SAS).

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q335 marksLong AnswerTriangles CBSE 2025 · Standard 30/4/1

The corresponding sides of $△ A B C$ and $△ P QR$ are in the ratio 3 : 5. $A D ⊥ B C$ and $P S ⊥ QR$ as shown in the following figures :
(i) Prove that $△ A D C \sim △ P S R$
(ii) If $A D = 4$ cm, find the length of PS.
(iii) Using (ii) find ar $(△ A B C)$ : ar $(△ P QR)$

Show answer & solution

Answer: (i) Proved. (ii)

P S = \frac{20}{3}

cm (iii)

9 : 25

(i) $△ A B C \sim △ P QR$ (sides proportional), so $∠ C = ∠ R$ .
In $△ A D C$ and $△ P S R$ : $∠ A D C = ∠ P S R = 9 0^{\circ}$ and $∠ C = ∠ R$ , so $△ A D C \sim △ P S R$ (AA).
(ii) $\frac{A D}{P S} = \frac{A C}{P R} = \frac{3}{5}$ , so $P S = \frac{5 \times 4}{3} = \frac{20}{3}$ cm
(iii) $\frac{ar ( △ A B C )}{ar ( △ P QR )} = \frac{\frac{1}{2} \times B C \times A D}{\frac{1}{2} \times QR \times P S} = \frac{3}{5} \times \frac{4}{20/3} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
So ar $(△ A B C)$ : ar $(△ P QR) = 9 : 25$

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2025 · Standard 30/4/1

State basic proportionality theorem. Use it to prove the following :
If three parallel lines $l$ , $m$ , $n$ are intersected by transversals $q$ and $s$ as shown in the adjoining figure, then $\frac{A B}{B C} = \frac{D E}{E F}$ .

Show answer & solution

Answer: Statement given; Proved.

BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof of BPT: In $△ A B C$ let $D E ∥ B C$ with D on AB, E on AC. Join BE and CD; draw $E N ⊥ A B$ , $D M ⊥ A C$ .
$\frac{ar ( A D E )}{ar ( B D E )} = \frac{\frac{1}{2} A D \cdot E N}{\frac{1}{2} D B \cdot E N} = \frac{A D}{D B}$ and $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$
$△ B D E$ and $△ D E C$ are on the same base DE between the same parallels DE and BC, so ar(BDE) = ar(DEC). Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .
Application: Join AF, cutting line $m$ at G.
In $△ A C F$ , $B G ∥ C F$ (as $m ∥ n$ ), so by BPT $\frac{A B}{B C} = \frac{A G}{GF}$ ... (1)
In $△ F A D$ , $GE ∥ A D$ (as $m ∥ l$ ), so by BPT $\frac{F G}{G A} = \frac{F E}{E D}$ , i.e. $\frac{A G}{GF} = \frac{D E}{E F}$ ... (2)
From (1) and (2): $\frac{A B}{B C} = \frac{D E}{E F}$

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q335 marksLong AnswerTriangles CBSE 2025 · Standard 30/5/1

If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points then it divides the two sides in the same ratio. Prove it.
Also, state the converse of the above statement.

Show answer & solution

Answer: Proved. Converse: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given: In $△ A B C$ , $D E ∥ B C$ , D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw $E M ⊥ A B$ and $D N ⊥ A C$ .
$ar (A D E) = \frac{1}{2} A D \cdot E M$ , $ar (B D E) = \frac{1}{2} D B \cdot E M$ , so $\frac{ar ( A D E )}{ar ( B D E )} = \frac{A D}{D B}$ .
$ar (A D E) = \frac{1}{2} A E \cdot D N$ , $ar (D E C) = \frac{1}{2} E C \cdot D N$ , so $\frac{ar ( A D E )}{ar ( D E C )} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so $ar (B D E) = ar (D E C)$ .
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .
Converse: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Q345 marksLong AnswerTriangles CBSE 2025 · Standard 30/5/2

State the converse of basic proportionality theorem.
Also find $\frac{B F}{F C}$ in the following figure, given that $A B ∥ D C ∥ E F$ and $\frac{A E}{E D} = \frac{2}{3}$ .
Also, find the length of EF if AB = 10 cm and DC = 15 cm.

Show answer & solution

Answer: Converse: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

\frac{B F}{F C} = \frac{2}{3}

; EF = 12 cm

Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Let diagonal AC meet EF at G.
In $△ A D C$ , $E G ∥ D C$ : $\frac{A G}{GC} = \frac{A E}{E D} = \frac{2}{3}$ (BPT)
In $△ C A B$ , $GF ∥ A B$ : $\frac{B F}{F C} = \frac{A G}{GC} = \frac{2}{3}$
$△ A E G \sim △ A D C$ : $\frac{E G}{D C} = \frac{A E}{A D} = \frac{2}{5} \Rightarrow E G = \frac{2}{5} \times 15 = 6$ cm
$△ C GF \sim △ C A B$ : $\frac{GF}{A B} = \frac{C G}{C A} = \frac{3}{5} \Rightarrow GF = \frac{3}{5} \times 10 = 6$ cm
$E F = E G + GF = 12$ cm

Q325 marksLong AnswerTriangles CBSE 2025 · Standard 30/5/3

State the basic proportionality theorem.
Use the theorem to do the following :
In $△ A B C$ , AD is the angle bisector of angle A. BA is produced to E such that $C E ∥ A D$ . Prove that $\frac{B D}{D C} = \frac{B A}{A C}$ .

Show answer & solution

Answer: BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

\frac{B D}{D C} = \frac{B A}{A C}

proved.

BPT: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
AD bisects $∠ A$ : $∠1 = ∠2$ (where $∠1 = ∠ B A D$ , $∠2 = ∠ D A C$ ).
$C E ∥ A D$ with transversal AC: $∠2 = ∠3$ (alternate angles, $∠3 = ∠ A C E$ ).
$C E ∥ A D$ with transversal BE: $∠1 = ∠4$ (corresponding angles, $∠4 = ∠ A E C$ ).
So $∠3 = ∠4$ , hence $A E = A C$ (sides opposite equal angles in $△ A C E$ ).
In $△ B C E$ , $A D ∥ E C$ , so by BPT $\frac{B D}{D C} = \frac{B A}{A E}$ .
Since $A E = A C$ : $\frac{B D}{D C} = \frac{B A}{A C}$ . Proved.

Q335 marksLong AnswerTriangles CBSE 2025 · Standard 30/6/1

If a line drawn parallel to one side of triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to third side.
State and prove the converse of the above statement.

Show answer & solution

Answer: Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio (Basic Proportionality Theorem). Proved.

Statement (BPT): In $△ A B C$ , if $D E ∥ B C$ with D on AB and E on AC, then $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .
$a r (A D E) = \frac{1}{2} A D \cdot E N$ , $a r (B D E) = \frac{1}{2} D B \cdot E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$
$a r (A D E) = \frac{1}{2} A E \cdot D M$ , $a r (D E C) = \frac{1}{2} E C \cdot D M$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$
$△ B D E$ and $△ D E C$ are on the same base DE between the parallels DE and BC, so $a r (B D E) = a r (D E C)$ .
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2025 Standard 30/6/3

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2025 · Standard 30/6/1

In the adjoining figure, $△ C A B$ is a right triangle, right angled at A and $A D ⊥ B C$ . Prove that $△ A D B \sim △ C D A$ . Further, if $B C = 10$ cm and $C D = 2$ cm, find the length of AD.

Show answer & solution

Answer: Proved; AD = 4 cm

In $△ A D B$ and $△ C D A$ : $∠ A D B = ∠ C D A = 9 0^{\circ}$
$∠ D A B = 9 0^{\circ} - ∠ D A C$ and, in $△ A D C$ , $∠ D C A = 9 0^{\circ} - ∠ D A C$ , so $∠ D A B = ∠ D C A$
Hence $△ A D B \sim △ C D A$ (AA similarity).
So $\frac{A D}{C D} = \frac{D B}{D A} \Rightarrow A D^{2} = C D \cdot D B$
$D B = B C - C D = 10 - 2 = 8$ cm
$A D^{2} = 2 \times 8 = 16 \Rightarrow A D = 4$ cm

Also asked in: 2025 Standard 30/6/3

Q345 marksLong AnswerTriangles CBSE 2024 · Basic 430/1/2

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Given: in $△ A B C$ , a line $D E ∥ B C$ meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .
ar(ADE) $= \frac{1}{2} \times A D \times E N$ and ar(BDE) $= \frac{1}{2} \times D B \times E N$ , so $\frac{ar(ADE)}{ar(BDE)} = \frac{A D}{D B}$ .
Similarly ar(ADE) $= \frac{1}{2} \times A E \times D M$ and ar(DEC) $= \frac{1}{2} \times E C \times D M$ , so $\frac{ar(ADE)}{ar(DEC)} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2024 Basic 430/1/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Basic 430/1/2

Sides AB and BC and median AD of a $△ A B C$ are respectively proportional to sides PQ and PR and median PM of $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Diagram for CBSE 2024 Class 10 Maths question 34 (OR)

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ , where AD and PM are medians.
Produce AD to E so that DE = AD, and PM to N so that MN = PM. Join BE, CE, QN and RN.
The diagonals of ABEC bisect each other at D, so ABEC is a parallelogram and BE = AC; similarly QN = PR.
Then $\frac{A B}{P Q} = \frac{B E}{QN} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ , so $△ A B E \sim △ P QN$ (SSS) and $∠ B A E = ∠ QP N$ .
In the same way $△ A C E \sim △ P R N$ , so $∠ C A E = ∠ R P N$ .
Adding, $∠ B A C = ∠ QP R$ .
With $\frac{A B}{P Q} = \frac{A C}{P R}$ , $△ A B C \sim △ P QR$ (SAS).

Also asked in: 2024 Basic 430/1/3

Q345 marksLong AnswerTriangles CBSE 2024 · Basic 430/2/1

In the given figure, altitudes CE and AD of $△ A B C$ intersect each other at the point P.
Show that
(i) $△ A E P \sim △ C D P$
(ii) $△ A B D \sim △ C B E$
(iii) $△ A E P \sim △ A D B$

Diagram for CBSE 2024 Class 10 Maths question 34

Show answer & solution

Answer: Proved.

(i) In $△ A E P$ and $△ C D P$ : $∠ A E P = ∠ C D P = 9 0^{\circ}$ and $∠ A P E = ∠ C P D$ (vertically opposite). So $△ A E P \sim △ C D P$ (AA).
(ii) In $△ A B D$ and $△ C B E$ : $∠ A D B = ∠ C E B = 9 0^{\circ}$ and $∠ A B D = ∠ C B E$ (common angle B). So $△ A B D \sim △ C B E$ (AA).
(iii) In $△ A E P$ and $△ A D B$ : $∠ A E P = ∠ A D B = 9 0^{\circ}$ and $∠ P A E = ∠ B A D$ (common angle A). So $△ A E P \sim △ A D B$ (AA).

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Basic 430/2/1

AD and PM are medians of triangles ABC and PQR, respectively, where $△ A B C \sim △ P QR$ . Prove that $\frac{A B}{P Q} = \frac{A D}{P M}$ .

Show answer & solution

Answer: Proved.

Since $△ A B C \sim △ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ .
D and M are midpoints, so BC = 2BD and QR = 2QM.
Hence $\frac{A B}{P Q} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$ .
In $△ A B D$ and $△ P QM$ : $\frac{A B}{P Q} = \frac{B D}{QM}$ and $∠ B = ∠ Q$ , so $△ A B D \sim △ P QM$ (SAS).
Therefore $\frac{A B}{P Q} = \frac{A D}{P M}$ .

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q345 marksLong AnswerTriangles CBSE 2024 · Basic 430/4/1

If BD and QM are medians of triangles ABC and PQR, respectively, where $△ A B C \sim △ P QR$ , prove that $\frac{A B}{P Q} = \frac{B D}{QM}$ .

Show answer & solution

Answer: Proved.

$△ A B C \sim △ P QR$ gives $\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ A = ∠ P$ .
D and M are mid-points of AC and PR, so $A D = \frac{1}{2} A C$ and $P M = \frac{1}{2} P R$ .
Hence $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{2 A D}{2 P M} = \frac{A D}{P M}$ .
In $△ A B D$ and $△ P QM$ : $\frac{A B}{P Q} = \frac{A D}{P M}$ and $∠ A = ∠ P$ .
So $△ A B D \sim △ P QM$ (SAS similarity).
Therefore $\frac{A B}{P Q} = \frac{B D}{QM}$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Basic 430/4/1

CD and GH are respectively the bisectors of $∠ A C B$ and $∠ E GF$ such that D and H lie on sides AB and FE of $△ A B C$ and $△ F E G$ respectively. If $△ A B C \sim △ F E G$ , show that :
(i) $\frac{C D}{G H} = \frac{A C}{F G}$
(ii) $△ D C B \sim △ H GE$

Show answer & solution

Answer: Proved.

$△ A B C \sim △ F E G$ gives $∠ A = ∠ F$ , $∠ B = ∠ E$ , $∠ A C B = ∠ F GE$ .
Halving, $∠ A C D = ∠ F G H$ and $∠ D C B = ∠ H GE$ .
(i) In $△ A C D$ and $△ F G H$ : $∠ A = ∠ F$ , $∠ A C D = ∠ F G H$ , so $△ A C D \sim △ F G H$ (AA).
Hence $\frac{C D}{G H} = \frac{A C}{F G}$ .
(ii) In $△ D C B$ and $△ H GE$ : $∠ B = ∠ E$ , $∠ D C B = ∠ H GE$ .
So $△ D C B \sim △ H GE$ (AA similarity).

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q345 marksLong AnswerTriangles CBSE 2024 · Basic 430/5/1

Sides AB, BC and median AD of $△ A B C$ are respectively proportional to sides PQ, QR and median PM of $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ , where D and M are mid-points of BC and QR.
Since $B D = \frac{1}{2} B C$ and $QM = \frac{1}{2} QR$ , $\frac{B D}{QM} = \frac{B C}{QR}$ .
So $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ , hence $△ A B D \sim △ P QM$ (SSS).
So $∠ B = ∠ Q$ .
In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ .
Hence $△ A B C \sim △ P QR$ (SAS).

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Basic 430/5/1

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Let DE $∥$ BC in $△ A B C$ with D on AB and E on AC. To prove $\frac{A D}{D B} = \frac{A E}{E C}$ .
Join BE and CD; draw EN $⊥$ AB and DM $⊥$ AC.
ar(ADE) $= \frac{1}{2} A D \cdot E N$ and ar(BDE) $= \frac{1}{2} D B \cdot E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
Similarly ar(ADE) $= \frac{1}{2} A E \cdot D M$ and ar(DEC) $= \frac{1}{2} E C \cdot D M$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
Triangles BDE and DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Standard 30/1/1

In the given figure PA, QB and RC are each perpendicular to AC. If AP = $x$ , BQ = $y$ and CR = $z$ , then prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$

Show answer & solution

Answer: Proved.

In $△ A B Q$ and $△ A C R$ : $∠ A$ is common and $∠ A B Q = ∠ A C R = 9 0^{\circ}$ , so $△ A B Q \sim △ A C R$ (AA).
So $\frac{B Q}{C R} = \frac{A B}{A C}$ , i.e. $\frac{y}{z} = \frac{A B}{A C}$ ... (1)
In $△ C B Q$ and $△ C A P$ : $∠ C$ is common and $∠ C B Q = ∠ C A P = 9 0^{\circ}$ , so $△ C B Q \sim △ C A P$ (AA).
So $\frac{B Q}{A P} = \frac{B C}{A C}$ , i.e. $\frac{y}{x} = \frac{B C}{A C}$ ... (2)
Adding (1) and (2): $\frac{y}{z} + \frac{y}{x} = \frac{A B + B C}{A C} = \frac{A C}{A C} = 1$ .
Dividing by $y$ : $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Standard 30/2/1

Sides AB and AC and median AD to $△ A B C$ are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ , with D, M the mid-points of BC, QR.
Produce AD to E so that DE = AD, and PM to N so that MN = PM. Join BE, CE, QN, RN.
The diagonals of ABEC bisect each other, so ABEC is a parallelogram and BE = AC. Similarly PQNR is a parallelogram and QN = PR.
So $\frac{A B}{P Q} = \frac{B E}{QN} = \frac{2 A D}{2 P M} = \frac{A E}{P N}$ , hence $△ A B E \sim △ P QN$ (SSS) and $∠ B A E = ∠ QP N$ ... (1)
In the same way, $\frac{A C}{P R} = \frac{C E}{R N} = \frac{A E}{P N}$ gives $△ A C E \sim △ P R N$ (SSS) and $∠ C A E = ∠ R P N$ ... (2)
Adding (1) and (2): $∠ B A C = ∠ QP R$ .
With $\frac{A B}{P Q} = \frac{A C}{P R}$ , $△ A B C \sim △ P QR$ (SAS).

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q335 marksLong AnswerTriangles CBSE 2024 · Standard 30/3/1

State and prove Basic Proportionality theorem.

Show answer & solution

Answer: Proved.

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In $△ A B C$ , DE $∥$ BC, with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw EM $⊥$ AB and DN $⊥$ AC.
ar(ADE) = $\frac{1}{2} \times A D \times E M$ and ar(BDE) = $\frac{1}{2} \times D B \times E M$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
ar(ADE) = $\frac{1}{2} \times A E \times D N$ and ar(DEC) = $\frac{1}{2} \times E C \times D N$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
Triangles BDE and DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2026 Standard 30/1/1, 2026 Standard 30/1/2, 2026 Standard 30/1/3, 2025 Basic 430/2/3, 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q355 marksLong AnswerTriangles CBSE 2024 · Standard 30/3/3

In $△ A B C$ , if AD $⊥$ BC and $A D^{2} = B D \times D C$ , then prove that $∠ B A C = 9 0^{\circ}$ .

Show answer & solution

Answer: Proved.

$A D^{2} = B D \times D C \Rightarrow \frac{A D}{B D} = \frac{D C}{A D}$ .
In $△ A D B$ and $△ C D A$ : $∠ A D B = ∠ C D A = 9 0^{\circ}$ and $\frac{A D}{C D} = \frac{B D}{A D}$ .
So $△ A D B \sim △ C D A$ (SAS similarity).
Hence $∠ B A D = ∠ A C D$ and $∠ A B D = ∠ C A D$ .
$∠ B A C = ∠ B A D + ∠ C A D = ∠ C + ∠ B$ .
In $△ A B C$ : $∠ A + ∠ B + ∠ C = 18 0^{\circ} \Rightarrow 2∠ B A C = 18 0^{\circ}$ .
So $∠ B A C = 9 0^{\circ}$ .

Also asked in: 2025 Standard 30/1/1, 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Standard 30/4/1

Sides AB, BC and the median AD of $△ A B C$ are respectively proportional to sides PQ, QR and the median PM of another $△ P QR$ . Prove that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ , where D and M are mid-points of BC and QR.
BD = $\frac{1}{2}$ BC and QM = $\frac{1}{2}$ QR, so $\frac{B D}{QM} = \frac{B C}{QR}$ .
Hence $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ , so $△ A B D \sim △ P QM$ (SSS similarity).
Therefore $∠ B = ∠ Q$ .
In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ and the included angles $∠ B = ∠ Q$ .
By SAS similarity, $△ A B C \sim △ P QR$ .

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q345 marksLong AnswerTriangles CBSE 2024 · Standard 30/5/1

In the given figure, $△ F E C ≅ △ G D B$ and $∠1 = ∠2$ .
Prove that $△ A D E \sim △ A B C$ .

Show answer & solution

Answer: Proved.

From $△ F E C ≅ △ G D B$ , the corresponding sides give EC = DB (CPCT).
$∠1 = ∠2$ means $∠ A D E = ∠ A E D$ , so AD = AE (sides opposite equal angles).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ , and by the converse of BPT, DE $∥$ BC.
So $∠ A D E = ∠ A B C$ and $∠ A E D = ∠ A C B$ (corresponding angles), and $∠ A$ is common.
Therefore $△ A D E \sim △ A B C$ (AA similarity).

Also asked in: 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2024 · Standard 30/5/1

Sides AB and AC and median AD of a $△ A B C$ are respectively proportional to sides PQ and PR and median PM of another $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$ .
Produce AD to E with DE = AD and join EC; produce PM to N with MN = PM and join NR.
$△ A B D ≅ △ E C D$ (SAS: BD = DC, AD = DE, vertically opposite angles), so EC = AB; similarly NR = PQ.
Then $\frac{E C}{N R} = \frac{A C}{P R} = \frac{A E}{P N}$ (as AE = 2AD, PN = 2PM), so $△ A E C \sim △ P N R$ (SSS) and $∠ C A E = ∠ R P N$ .
Similarly, using $△ A E B$ (BE = AC) and $△ P N Q$ , we get $∠ B A E = ∠ QP N$ .
Adding, $∠ B A C = ∠ QP R$ . With $\frac{A B}{P Q} = \frac{A C}{P R}$ , $△ A B C \sim △ P QR$ (SAS similarity).

Also asked in: 2024 Standard 30/5/2, 2024 Standard 30/5/3

Q345 marksLong AnswerTriangles CBSE 2023 · Basic 430/5/1

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then prove that the other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Given: In $△ A B C$ , DE $∥$ BC meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw EM $⊥$ AB and DN $⊥$ AC.
ar(ADE) $= \frac{1}{2} \times A D \times E M$ and ar(BDE) $= \frac{1}{2} \times D B \times E M$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
Similarly ar(ADE) $= \frac{1}{2} \times A E \times D N$ and ar(DEC) $= \frac{1}{2} \times E C \times D N$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ . Proved.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q355 marksLong AnswerTriangles CBSE 2023 · Basic 430/6/1

Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points, divides the two sides in the same ratio.

Show answer & solution

Answer: Proved.

Given: In $△ A B C$ , a line DE $∥$ BC meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw EN $⊥$ AB and DM $⊥$ AC.
ar(ADE) $= \frac{1}{2} \times A D \times E N$ and ar(BDE) $= \frac{1}{2} \times D B \times E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
Similarly ar(ADE) $= \frac{1}{2} \times A E \times D M$ and ar(DEC) $= \frac{1}{2} \times E C \times D M$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
Triangles BDE and DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Therefore $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q35 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Basic 430/6/1

In the given figure, $\frac{QR}{QS} = \frac{QT}{P R}$ and $∠1 = ∠2$ . Prove that $△ P QS \sim △ T QR$ .

Diagram for CBSE 2023 Class 10 Maths question 35 (OR)

Show answer & solution

Answer: Proved.

In $△ P QR$ , $∠1 = ∠2$ i.e. $∠ P QR = ∠ P R Q$ , so $P Q = P R$ (sides opposite equal angles).
Given $\frac{QR}{QS} = \frac{QT}{P R} = \frac{QT}{P Q}$ , so $\frac{QS}{QR} = \frac{P Q}{QT}$ .
In $△ P QS$ and $△ T QR$ : $\frac{P Q}{T Q} = \frac{QS}{QR}$ and $∠ P QS = ∠ T QR$ (common angle $∠1$ ).
By SAS similarity, $△ P QS \sim △ T QR$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/3/3, 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q325 marksLong AnswerTriangles CBSE 2023 · Standard 30/1/1

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, prove that the other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Given: In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .
$\frac{a r ( A D E )}{a r ( B D E )} = \frac{\frac{1}{2} A D \cdot E N}{\frac{1}{2} D B \cdot E N} = \frac{A D}{D B}$ .
$\frac{a r ( A D E )}{a r ( D E C )} = \frac{\frac{1}{2} A E \cdot D M}{\frac{1}{2} E C \cdot D M} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Q345 marksLong AnswerTriangles CBSE 2023 · Standard 30/1/2

PA, QB and RC are each perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, then prove that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ .

Diagram for CBSE 2023 Class 10 Maths question 34

Show answer & solution

Answer: Proved.

In $△ A B Q$ and $△ A C R$ : $∠ A$ is common and $∠ A B Q = ∠ A C R = 9 0^{\circ}$ , so $△ A B Q \sim △ A C R$ (AA).
$\frac{QB}{R C} = \frac{A B}{A C}$ , i.e. $\frac{z}{y} = \frac{a}{a + b}$ ... (1)
In $△ C B Q$ and $△ C A P$ : $∠ C$ is common and $∠ C B Q = ∠ C A P = 9 0^{\circ}$ , so $△ C B Q \sim △ C A P$ (AA).
$\frac{QB}{P A} = \frac{C B}{C A}$ , i.e. $\frac{z}{x} = \frac{b}{a + b}$ ... (2)
Adding (1) and (2): $\frac{z}{x} + \frac{z}{y} = \frac{a + b}{a + b} = 1$ .
Dividing by z: $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ .

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/1/2

In the given figure, CD and RS are respectively the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ then prove that :
(i) $△ A D C \sim △ P S R$
(ii) $A D \times P R = A C \times P S$

Diagram for CBSE 2023 Class 10 Maths question 34 (OR)

Show answer & solution

Answer: Proved.

$△ A B C \sim △ P QR$ , so $∠ A = ∠ P$ and $\frac{A B}{P Q} = \frac{A C}{P R}$ .
D and S are mid-points of AB and PQ, so $A B = 2 A D$ and $P Q = 2 P S$ ; hence $\frac{A D}{P S} = \frac{A B}{P Q} = \frac{A C}{P R}$ .
(i) In $△ A D C$ and $△ P S R$ : $\frac{A D}{P S} = \frac{A C}{P R}$ and $∠ A = ∠ P$ , so $△ A D C \sim △ P S R$ (SAS).
(ii) From (i), $\frac{A D}{P S} = \frac{A C}{P R}$ , so $A D \times P R = A C \times P S$ .

Q345 marksLong AnswerTriangles CBSE 2023 · Standard 30/1/3

Sides AB and AC and median AM of a $△ A B C$ are proportional to sides DE and DF and Median DN of another $△ D E F$ . Show that $△ A B C \sim △ D E F$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{D E} = \frac{A C}{D F} = \frac{A M}{D N}$ . Produce AM to P with MP = AM, and DN to Q with NQ = DN; join CP and FQ.
In $△ A B M$ and $△ P C M$ : BM = CM, AM = PM, $∠ A M B = ∠ P M C$ , so they are congruent (SAS) and CP = AB. Similarly FQ = DE.
So $\frac{C P}{F Q} = \frac{A C}{D F} = \frac{A P}{D Q}$ (as AP = 2AM, DQ = 2DN). Hence $△ A C P \sim △ D F Q$ (SSS) and $∠ C A P = ∠ F D Q$ .
In the same way (using BP = AC, EQ = DF) $△ A B P \sim △ D E Q$ , so $∠ B A P = ∠ E D Q$ .
Adding, $∠ B A C = ∠ E D F$ .
In $△ A B C$ and $△ D E F$ : $\frac{A B}{D E} = \frac{A C}{D F}$ and $∠ A = ∠ D$ , so $△ A B C \sim △ D E F$ (SAS).

Q34 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/1/3

ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at L. Prove that
(i) $\frac{D P}{P L} = \frac{D C}{B L}$
(ii) $\frac{D L}{D P} = \frac{A L}{D C}$
(iii) If LP : PD = 2 : 3 then find BP : BC.

Show answer & solution

Answer: (i) Proved. (ii) Proved. (iii) BP : BC = 2 : 5

(i) In $△ D P C$ and $△ L P B$ : $∠ D P C = ∠ L P B$ (vertically opposite) and $∠ D C P = ∠ L B P$ (alternate angles, $D C ∥ A L$ ). So $△ D P C \sim △ L P B$ (AA), giving $\frac{D P}{P L} = \frac{D C}{B L}$ .
(ii) In $△ A L D$ and $△ C D P$ : $∠ A L D = ∠ C D P$ (alternate angles, $A L ∥ D C$ ) and $∠ D A L = ∠ P C D$ (opposite angles of a parallelogram). So $△ A L D \sim △ C D P$ (AA), giving $\frac{D L}{D P} = \frac{A L}{C D} = \frac{A L}{D C}$ .
(iii) From $△ D P C \sim △ L P B$ : $\frac{B P}{P C} = \frac{L P}{P D} = \frac{2}{3}$ .
So BP = 2k, PC = 3k, BC = 5k, and BP : BC = 2 : 5.

Q355 marksLong AnswerTriangles CBSE 2023 · Standard 30/2/1

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $△ P QR$ . Show that $△ A B C \sim △ P QR$ .

Show answer & solution

Answer: Proved.

Given $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A D}{P M}$ , with D and M midpoints of BC and QR.
$B D = \frac{1}{2} B C$ and $QM = \frac{1}{2} QR$ , so $\frac{B D}{QM} = \frac{B C}{QR}$ .
Hence $\frac{A B}{P Q} = \frac{B D}{QM} = \frac{A D}{P M}$ , so $△ A B D \sim △ P QM$ (SSS).
Therefore $∠ B = ∠ Q$ .
In $△ A B C$ and $△ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ .
So $△ A B C \sim △ P QR$ (SAS).

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q35 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/2/1

Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E. Prove that EL = 2BL.

Show answer & solution

Answer: Proved.

In $△ B M C$ and $△ E M D$ : MC = MD, $∠ B M C = ∠ E M D$ (vertically opposite), $∠ B C M = ∠ E D M$ (alternate angles, $B C ∥ A E$ ).
So $△ B M C ≅ △ E M D$ (ASA), giving DE = BC.
AD = BC, so AE = AD + DE = 2BC.
In $△ A L E$ and $△ C L B$ : $∠ A L E = ∠ C L B$ (vertically opposite), $∠ L A E = ∠ L C B$ (alternate angles).
So $△ A L E \sim △ C L B$ (AA) and $\frac{E L}{B L} = \frac{A E}{C B} = 2$ .
Hence EL = 2BL.

Also asked in: 2024 Standard 30/3/2, 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q335 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/1

D is a point on the side BC of a triangle ABC such that $∠ A D C = ∠ B A C$ , prove that $C A^{2} = C B . C D$

Show answer & solution

Answer: Proved.

In $△ A D C$ and $△ B A C$ :
$∠ A D C = ∠ B A C$ (given)
$∠ A C D = ∠ B C A$ (common angle C)
So $△ A D C \sim △ B A C$ (AA similarity).
Corresponding sides are proportional: $\frac{C A}{C B} = \frac{C D}{C A}$
Hence $C A^{2} = C B \cdot C D$ .

Q33 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/1

If AD and PM are medians of triangles ABC and PQR, respectively where $△ A B C \sim △ P QR$ , prove that $\frac{A B}{P Q} = \frac{A D}{P M}$ .

Show answer & solution

Answer: Proved.

Since $△ A B C \sim △ P QR$ : $\frac{A B}{P Q} = \frac{B C}{QR}$ and $∠ B = ∠ Q$ .
D and M are mid-points of BC and QR, so $B D = \frac{1}{2} B C$ and $QM = \frac{1}{2} QR$ .
Hence $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{2 B D}{2 QM} = \frac{B D}{QM}$ .
In $△ A B D$ and $△ P QM$ : $\frac{A B}{P Q} = \frac{B D}{QM}$ and $∠ B = ∠ Q$ , so $△ A B D \sim △ P QM$ (SAS similarity).
Therefore $\frac{A B}{P Q} = \frac{A D}{P M}$ .

Q355 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/2

In the given figure, $∠ A D C = ∠ B C A$ ; prove that $△ A C B \sim △ A D C$ . Hence find BD if AC = 8 cm and AD = 3 cm.

Diagram for CBSE 2023 Class 10 Maths question 35

Show answer & solution

Answer: BD

= \frac{55}{3}

cm ≈ 18.33 cm

In $△ A C B$ and $△ A D C$ :
$∠ B C A = ∠ A D C$ (given) and $∠ C A B = ∠ D A C$ (common angle A)
So $△ A C B \sim △ A D C$ (AA similarity).
Hence $\frac{A C}{A D} = \frac{A B}{A C}$ , so $A B = \frac{A C ^{2}}{A D} = \frac{64}{3}$ cm
$B D = A B - A D = \frac{64}{3} - 3 = \frac{55}{3}$ cm

Q35 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/2

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Show answer & solution

Answer: Proved.

Given: In $△ A B C$ , a line DE $∥$ BC meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: join BE and CD; draw EM $⊥$ AB and DN $⊥$ AC.
ar(ADE) $= \frac{1}{2} A D \cdot E M$ , ar(BDE) $= \frac{1}{2} D B \cdot E M$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
ar(ADE) $= \frac{1}{2} A E \cdot D N$ , ar(DEC) $= \frac{1}{2} E C \cdot D N$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Therefore $\frac{A D}{D B} = \frac{A E}{E C}$ .

Q325 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/3

In a $△ P QR$ , N is a point on PR, such that QN $⊥$ PR. If PN × NR = $Q N^{2}$ , prove that $∠ P QR = 9 0^{\circ}$ .

Show answer & solution

Answer: Proved.

$P N \times N R = Q N^{2} \Rightarrow \frac{P N}{QN} = \frac{QN}{N R}$
In $△ P N Q$ and $△ QN R$ : $\frac{P N}{QN} = \frac{QN}{N R}$ and $∠ P N Q = ∠ QN R = 9 0^{\circ}$ .
So $△ P N Q \sim △ QN R$ (SAS similarity), giving $∠ QP N = ∠ R QN$ .
In right $△ P N Q$ : $∠ QP N + ∠ P QN = 9 0^{\circ}$ .
So $∠ R QN + ∠ P QN = 9 0^{\circ}$ , i.e. $∠ P QR = 9 0^{\circ}$ .

Q32 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/3

In the given figure, $△ A B C$ and $△ D B C$ are on the same base BC. If AD intersects BC at O, prove that $\frac{a r ( △ A B C )}{a r ( △ D B C )} = \frac{A O}{D O}$

Diagram for CBSE 2023 Class 10 Maths question 32 (OR)

Show answer & solution

Answer: Proved.

Draw AM $⊥$ BC and DN $⊥$ BC (M, N on line BC).
$\frac{a r ( △ A B C )}{a r ( △ D B C )} = \frac{\frac{1}{2} B C \cdot A M}{\frac{1}{2} B C \cdot D N} = \frac{A M}{D N}$
In $△ A M O$ and $△ D N O$ : $∠ A M O = ∠ D N O = 9 0^{\circ}$ and $∠ A O M = ∠ D O N$ (vertically opposite).
So $△ A M O \sim △ D N O$ (AA similarity), giving $\frac{A M}{D N} = \frac{A O}{D O}$ .
Hence $\frac{a r ( △ A B C )}{a r ( △ D B C )} = \frac{A O}{D O}$ .

← Arithmetic Progressions Coordinate Geometry →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →