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

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/3/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2025: question 33(b)

Set in the CBSE Class 10 board exam of 2025, 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 $Δ A B C$ and $Δ P QR$ . If $Δ A B C \sim Δ P QR$ , prove that:

(i) $Δ A M C \sim Δ P N R$

(ii) $∠ B C M = ∠ QR N$

(iii) $Δ B M C \sim Δ QN R$

Figure in the paper: Triangle ABC with M the midpoint of AB (M lies on side AB) and median CM drawn from C. Triangle PQR with N the midpoint of PQ (N lies on side QP) and median RN drawn from R. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/3/1.

Hints

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

Hint 1

A median cuts the side in half, so AM = ½AB and PN = ½PQ. How do AM/PN and AB/PQ compare?

Hint 2

For (ii), use the angle at C in part (i) and the whole angle C of the big triangles; for (iii), repeat the method of (i) using angle B.

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.

(i) Since $Δ A B C \sim Δ P QR$ , $\frac{A B}{P Q} = \frac{A C}{P R}$ .
½ mark
(i) M and N are midpoints of AB and PQ, so $A M = \frac{1}{2} A B$ and $P N = \frac{1}{2} P Q$ . Then $\frac{A M}{P N} = \frac{\frac{1}{2} A B}{\frac{1}{2} P Q} = \frac{A B}{P Q} = \frac{A C}{P R}$ .
½ mark
(i) Also $∠ A = ∠ P$ [corresponding angles of similar triangles].
½ mark
(i) Hence $Δ A M C \sim Δ P N R$ [SAS similarity criterion].
½ mark
(ii) From (i), $∠ A C M = ∠ P R N$ [corresponding angles]. Also $∠ A C B = ∠ P R Q$ [as $Δ A B C \sim Δ P QR$ ].
½ mark
(ii) Subtracting: $∠ A C B - ∠ A C M = ∠ P R Q - ∠ P R N$ , so $∠ B C M = ∠ QR N$ .
½ mark
(iii) Since $Δ A B C \sim Δ P QR$ , $\frac{A B}{P Q} = \frac{B C}{QR}$ .
½ mark
(iii) $B M = \frac{1}{2} A B$ and $QN = \frac{1}{2} P Q$ , so $\frac{B M}{QN} = \frac{A B}{P Q} = \frac{B C}{QR}$ .
½ mark
(iii) Also $∠ B = ∠ Q$ [corresponding angles of similar triangles].
½ mark
(iii) Hence $Δ B M C \sim Δ QN R$ [SAS similarity criterion].
½ mark

Answer: (i) ΔAMC ~ ΔPNR (SAS); (ii) ∠BCM = ∠QRN; (iii) ΔBMC ~ ΔQNR (SAS)

Where the marks are

(i) Writes AB/PQ = AC/PR from the given similarity (½ mark)
(i) Uses the medians to get AM/PN = AC/PR (½ mark)
(i) States ∠A = ∠P (½ mark)
(i) Concludes ΔAMC ~ ΔPNR by SAS (½ mark)
(ii) Gets ∠ACM = ∠PRN and ∠ACB = ∠PRQ (½ mark)
(ii) Subtracts to get ∠BCM = ∠QRN (½ mark)
(iii) Writes AB/PQ = BC/QR (½ mark)
(iii) Gets BM/QN = BC/QR (½ mark)
(iii) States ∠B = ∠Q (½ mark)
(iii) Concludes ΔBMC ~ ΔQNR by SAS (½ 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 paper2025 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 2025: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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