(B) In the given figure, CM and RN are respectively the medians of △ABC and △PQR. If $\triangle ABC \sim \triangle PQR$ , then prove that : (i) $\triangle AMC \sim \triangle PNR$ (ii) $\triangle CMB \sim \triangle RNQ$ .

Q: 34. (B) In the given figure, CM and RN are respectively the medians of △ABC and △PQR. If $\triangle ABC \sim \triangle PQR$, then prove that : (i) $\triangle AMC \sim \triangle PNR$ (ii) $\triangle CMB \sim \triangle RNQ$.

Since $\triangle ABC \sim \triangle PQR$: $AB/PQ = BC/QR = CA/RP$ and $\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R$. **(i) $\triangle AMC \sim \triangle PNR$:** CM, RN are medians ⇒ M, N are midpoints of AB, PQ ⇒ $AM = \frac{1}{2}AB, PN = \frac{1}{2}PQ$ ⇒ $AM/PN = AB/PQ = CA/RP$. Also $\angle A = \angle P$. By SAS similarity, $\triangle AMC \sim \triangle PNR$. **(ii) $\triangle CMB \sim \triangle RNQ$:** Similarly $MB = \frac{1}{2}AB, NQ = \frac{1}{2}PQ$ ⇒ $MB/NQ = AB/PQ = BC/QR$. Also $\angle B = \angle Q$. By SAS similarity, $\triangle CMB \sim \triangle RNQ$.

Triangles

Similar triangles and their medians

Official Answer

Since $\triangle ABC \sim \triangle PQR$ : $AB/PQ = BC/QR = CA/RP$ and $\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R$ .

(i) $\triangle AMC \sim \triangle PNR$ : CM, RN are medians ⇒ M, N are midpoints of AB, PQ ⇒ $AM = \frac{1}{2}AB, PN = \frac{1}{2}PQ$ ⇒ $AM/PN = AB/PQ = CA/RP$ . Also $\angle A = \angle P$ . By SAS similarity, $\triangle AMC \sim \triangle PNR$ .

(ii) $\triangle CMB \sim \triangle RNQ$ : Similarly $MB = \frac{1}{2}AB, NQ = \frac{1}{2}PQ$ ⇒ $MB/NQ = AB/PQ = BC/QR$ . Also $\angle B = \angle Q$ . By SAS similarity, $\triangle CMB \sim \triangle RNQ$ .

similar trianglesSAS similaritymedian bisects sideAM = half ABcorresponding angles equalAB/PQ = AC/PRratio of medianscorrespondence of vertices

Marking Scheme

11 mark: writing the consequences of $\triangle ABC \sim \triangle PQR$ — proportional sides $AB/PQ = BC/QR = CA/RP$ and equal angles $\angle A = \angle P, \angle B = \angle Q$ .
21 mark: using the median property to get $AM = \frac{1}{2}AB, PN = \frac{1}{2}PQ$ (and $MB = \frac{1}{2}AB, NQ = \frac{1}{2}PQ$ ).
31 mark (part i): showing $AM/PN = AC/PR$ and $\angle A = \angle P$ , hence $\triangle AMC \sim \triangle PNR$ by SAS similarity.
41 mark (part ii): showing $MB/NQ = BC/QR$ and $\angle B = \angle Q$ , hence $\triangle CMB \sim \triangle RNQ$ by SAS similarity.
51 mark: correctly naming the SAS similarity criterion in both parts and writing correspondence in proper order (accept the median-ratio corollary $CM/RN = AB/PQ$ as bonus reasoning).

Hint

A median bisects a side, so $AM = \frac{1}{2}AB$ and $PN = \frac{1}{2}PQ$ ; form $AM/PN$ , simplify to $AB/PQ = AC/PR$ , pair it with the equal included angle $\angle A = \angle P$ , and apply SAS similarity (repeat at B and Q for part ii).

Quick Oral Answer

Because $\triangle ABC \sim \triangle PQR$ the sides are proportional and angles equal, and since a median bisects a side, $AM/PN = \frac{\frac{1}{2}AB}{\frac{1}{2}PQ} = AB/PQ = AC/PR$ with $\angle A = \angle P$ , so $\triangle AMC \sim \triangle PNR$ by SAS; the same argument at B and Q gives $\triangle CMB \sim \triangle RNQ$ .

Analysis & Explanation

A SAS-similarity proof combining the median property with the given similarity of the parent triangles.

Concept

A median bisects its side, so $AM = MB = \frac{1}{2}AB$ and $PN = NQ = \frac{1}{2}PQ$ ; the ½ factors cancel when forming ratios like AM/PN, leaving AB/PQ.
Since △ABC ~ △PQR gives $AB/PQ = CA/RP = BC/QR$ , this re-expresses the median-half ratio as the side ratio needed for SAS: $AM/PN = CA/RP$ (part i) and $MB/NQ = BC/QR$ (part ii).
Combined with the equal included angle ( $\angle A = \angle P$ for part i; $\angle B = \angle Q$ for part ii), SAS similarity applies directly.

Common Mistakes

Trying to force AA similarity by hunting for an extra equal angle that isn't given, instead of using the intended SAS route.
Mismatching the sub-triangle correspondence: the relevant angle for △AMC is at A, but for △CMB it is at B.

Real-World / Extension

A useful corollary: the ratio of corresponding medians of similar triangles equals the ratio of corresponding sides ( $CM/RN = AB/PQ$ ); this extends to altitudes and angle bisectors and often appears as a separate 1–2 mark result.

Common Mistakes

1Trying to use AA similarity by assuming a second equal angle that is not given, instead of using the intended SAS criterion with the median-halved sides.
2Pairing the wrong included angle — using ∠A for △CMB instead of ∠B — which breaks the SAS correspondence.
3Forgetting to state that a median bisects the side ( $AM = \frac{1}{2}AB, PN = \frac{1}{2}PQ$ ), which is the key step that lets the ½ factors cancel to give $AB/PQ$ .

Interesting Facts

This proof leads to a widely used result: in similar triangles, the ratio of corresponding medians equals the ratio of corresponding sides — the same is true for corresponding altitudes and angle bisectors, all sharing the single scale factor of the similarity.

Because areas of similar triangles scale as the square of the side ratio, the medians (which scale linearly like the sides) give a quick way to find the area ratio: $(CM/RN)^2 = \frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)}$ .

SAS similarity requires only two pairs of proportional sides and the included angle, making it the most economical of the three similarity criteria — perfectly suited to problems like this where a bisected side gives one ratio for free.

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Frequently Asked Questions

Which similarity criterion is used here and why?

The SAS (Side–Angle–Side) similarity criterion is used. We show two pairs of corresponding sides are proportional (e.g. $AM/PN = AC/PR$ ) and the included angles are equal ( $\angle A = \angle P$ ). SAS is chosen because the median gives us one side ratio directly and the parent similarity gives the included angle — no need to find a third side or a second angle.

Why does AM/PN simplify to AB/PQ?

Because CM and RN are medians, M is the midpoint of AB and N is the midpoint of PQ. So $AM = \frac{1}{2}AB$ and $PN = \frac{1}{2}PQ$ . When you form $AM/PN = \frac{\frac{1}{2}AB}{\frac{1}{2}PQ}$ , the halves cancel, leaving $AB/PQ$ , which by the given similarity equals AC/PR and BC/QR.

What useful result follows from this proof?

It follows that the ratio of the corresponding medians of two similar triangles equals the ratio of their corresponding sides: $CM/RN = AB/PQ$ . This is a standard corollary that also holds for corresponding altitudes and angle bisectors, and it is often asked directly in exams.