OR

In the given figure, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, then prove that :

(i) Δ AMC ~ Δ PNR

(ii) Δ CMB ~ Δ RNQ

Q: Why is SAS used here instead of AA similarity?

Only one pair of angles ($\angle A$ or $\angle B$) is directly known to be equal from the given similarity, along with a derivable side ratio from the median being half of the corresponding side, so SAS (two sides and the included angle) is the applicable and provable criterion, not AA.

Triangle ABC with C at the top apex, A bottom-left, B bottom-right, M is the midpoint of AB (median CM drawn from C to M — Fig. for Q40

Triangles

Similarity of Triangles using Medians

Official Answer

Both $\Delta AMC \sim \Delta PNR$ and $\Delta CMB \sim \Delta RNQ$ are proved by the SAS similarity criterion: from $\Delta ABC \sim \Delta PQR$ , corresponding angles are equal and sides proportional, and since M, N are midpoints of AB, PQ, $\frac{AM}{PN} = \frac{BM}{QN} = \frac{AB}{PQ}$ , giving the second proportional side for each triangle pair with the already-equal included angle.

similar trianglesSAS similarity criterionmedian of a trianglemidpointproportional sidescorresponding anglesDelta ABC similar Delta PQR

Marking Scheme

11 mark: correctly stating Given/To Prove and listing the equal angles and proportional sides from $\Delta ABC \sim \Delta PQR$ .
21 mark: correctly showing $\frac{AM}{PN} = \frac{AB}{PQ} = \frac{CA}{RP}$ using M, N as midpoints.
31 mark: applying SAS similarity criterion (with included angle $\angle A = \angle P$ ) to conclude $\Delta AMC \sim \Delta PNR$ .
41 mark: correctly showing $\frac{BM}{QN} = \frac{AB}{PQ} = \frac{BC}{QR}$ using M, N as midpoints.
51 mark: applying SAS similarity criterion (with included angle $\angle B = \angle Q$ ) to conclude $\Delta CMB \sim \Delta RNQ$ .

Hint

Use $AM = \frac{AB}{2}$ , $PN = \frac{PQ}{2}$ (and $BM = \frac{AB}{2}$ , $QN = \frac{PQ}{2}$ ) with the given similarity ratio $\frac{AB}{PQ} = \frac{CA}{RP} = \frac{BC}{QR}$ , then apply SAS similarity using the already-equal included angles $\angle A$ and $\angle B$ .

Quick Oral Answer

Since M and N are midpoints of AB and PQ, the median segments AM, PN and BM, QN take exactly half of AB and PQ respectively, so their ratio equals AB/PQ, which matches the already-given similarity ratios; combined with the equal included angles $\angle A$ and $\angle B$ from $\Delta ABC \sim \Delta PQR$ , SAS similarity directly gives $\Delta AMC \sim \Delta PNR$ and $\Delta CMB \sim \Delta RNQ$ .

Analysis & Explanation

This extends similarity to medians, applying the SAS similarity criterion to two derived triangles.

Key insight

Since M, N are midpoints, $AM = \frac{AB}{2}$ , $PN = \frac{PQ}{2}$ , so $\frac{AM}{PN} = \frac{AB}{PQ}$ , which already equals the known similarity ratio $\frac{CA}{RP}$ (or $\frac{BC}{QR}$ for the other pair).
The included angle ( $\angle A = \angle P$ , or $\angle B = \angle Q$ ) is inherited unchanged from ΔABC ~ ΔPQR since M, N lie on AB, PQ without altering these angles.

Common mistakes

Attempting AA similarity by guessing a second angle instead of the cleaner, intended SAS route using one known angle plus the derived median-side ratio.

Broader/real-world significance

Proves that corresponding medians of similar triangles are proportional to corresponding sides — a result reused for altitudes, angle bisectors, perimeters and areas of similar triangles.

Common Mistakes

1Trying to use AA similarity by inventing a second equal angle instead of correctly applying SAS with the derived side ratio.
2Forgetting to state that $AM = \frac{AB}{2}$ and $PN = \frac{PQ}{2}$ explicitly before equating the ratio to $\frac{AB}{PQ}$ .
3Mixing up which included angle ( $\angle A$ vs $\angle B$ ) goes with which pair of triangles in the two separate proofs.

Interesting Facts

This result generalises: for any two similar triangles, corresponding medians, altitudes, and angle bisectors are all in the same ratio as the corresponding sides — a property frequently used in area-ratio problems (ratio of areas = square of ratio of corresponding medians).

The median-based similarity proof is a standard NCERT exemplar problem and one of the most repeated 'OR' alternative questions in CBSE Board papers for the Triangles chapter.

This same SAS-based technique can be extended to prove that corresponding altitudes of similar triangles are proportional to corresponding sides, a related and equally common CBSE proof question.

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Frequently Asked Questions

Why is SAS used here instead of AA similarity?

Only one pair of angles ( $\angle A$ or $\angle B$ ) is directly known to be equal from the given similarity, along with a derivable side ratio from the median being half of the corresponding side, so SAS (two sides and the included angle) is the applicable and provable criterion, not AA.

Does this proof work only for medians, or for any cevian?

The same reasoning extends to altitudes and angle bisectors of similar triangles as well — any corresponding cevian from a similar triangle pair will be proportional to the corresponding sides, provable by an analogous SAS argument.

What is the practical use of proving corresponding medians are proportional?

It allows calculating unknown median lengths or ratios in one triangle if the corresponding median in a similar triangle is known, useful in problems involving centroids and area ratios of similar triangles.