In the given figure, CM and RN are respectively, the medians of △ ABC and △ PQR. If △ ABC ~ △ PQR, prove that: (i) △ AMC ~ △ PNR (ii) ∠ BCM = ∠ QRN (iii) △ BMC ~ △ QNR

(i) △ AMC ~ △ PNR (SAS); (ii) ∠ BCM = ∠ QRN; (iii) △ BMC ~ △ QNR (AA)

MathsCBSE Class 10 · Basic · Board exam 2026, set 430/3/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2026: question 33(b)

Set in the CBSE Class 10 board exam of 2026, paper 430/3/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerTriangles

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) $△ B M C \sim △ QN R$

Figure in the paper: Triangle ABC with M the midpoint of AB, CM joined; triangle PQR with N the midpoint of PQ, RN joined. Vertices: A top, B left, C bottom; P top right, Q left, R bottom. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/3/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write all the equal angles and equal ratios that follow from $△ A B C \sim △ P QR$ .

Hint 2

Since M and N are midpoints, AM and PN are halves of AB and PQ, so their ratio equals $\frac{A B}{P Q}$ .

Hint 3

For (ii), subtract $∠ A C M$ from $∠ A C B$ . For (iii), combine $∠ B = ∠ Q$ with the result of (ii).

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Since $△ A B C \sim △ P QR$ : $∠ A = ∠ P$ , $∠ B = ∠ Q$ , $∠ A C B = ∠ P R Q$ and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{A C}{P R}$ ...(1)
1 mark
(i) M and N are midpoints, so $A M = \frac{1}{2} A B$ and $P N = \frac{1}{2} P Q$ , giving $\frac{A M}{P N} = \frac{A B}{P Q} = \frac{A C}{P R}$ . In $△ A M C$ and $△ P N R$ we also have $∠ M A C = ∠ N P R$ . Therefore $△ A M C \sim △ P N R$ [SAS similarity].
1½ marks
(ii) From (i), $∠ A C M = ∠ P R N$ [corresponding angles of similar triangles]. Also $∠ A C B = ∠ P R Q$ from (1). Subtracting, $∠ A C B - ∠ A C M = ∠ P R Q - ∠ P R N$ , i.e. $∠ B C M = ∠ QR N$ .
1½ marks
(iii) In $△ B M C$ and $△ QN R$ : $∠ M B C = ∠ N QR$ (as $∠ B = ∠ Q$ ) and $∠ B C M = ∠ QR N$ (from (ii)). Therefore $△ B M C \sim △ QN R$ [AA similarity].
1 mark

Answer: (i) $△ A M C \sim △ P N R$ (SAS); (ii) $∠ B C M = ∠ QR N$ ; (iii) $△ B M C \sim △ QN R$ (AA)

Where the marks are

Writes the equal angles and proportional sides from the given similarity (1 mark)
Uses the medians to get $\frac{A M}{P N} = \frac{A C}{P R}$ and proves (i) by SAS (1½ marks)
Deduces $∠ B C M = ∠ QR N$ by subtracting equal angles (1½ marks)
Proves (iii) by AA (1 mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 430/3/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2026: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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