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In the given figure, CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ∼ ΔPQR, then prove that ΔAMC ∼ ΔPNR.

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Question

In the given figure, CM and RN are respectively the medians of ΔABC and ΔPQR. If ΔABC ∼ ΔPQR, then prove that ΔAMC ∼ ΔPNR.

Theorem
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Solution

Given: ΔABC ∼ ΔPQR

Their corresponding sides are proportional, and their corresponding angles are equal: 

∴ ∠ A = ∠ P, ∠ B = ∠ = Q, ∠ C = ∠ R    ...(1)

`(AB)/(PQ) = (BC)/(QR) = (AC)/(PR)`   ...(2)

CM and RN are medians of ΔABC and ΔPQR respectively.

In Δ AMC ∼ Δ PNR

∠A = ∠P     ...(Included angle is equal)

`(AB)/(PQ) = (AC)/(PR)`    ...[From equation (2)]

`(2AM)/(2PN) = (AC)/(PR)`  

`(AM)/(PN) = (AC)/(PR)`    ...(Sides are proportional)

Since two sides are proportional and the included angle is equal, the triangles are similar by the SAS (Side-Angle-Side) similarity criterion. 

∴ ΔAMC ∼ ΔPNR

Hence Proved.

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2022-2023 (March) Basic - Outside Delhi Set 1
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