# Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle PQR. Prove triangle ABC ~ triangle PQR.

The proof of the statement is based on similarity in triangles.

## Answer: If ΔABC and ΔPQR in which AB, BC, and median AD of ΔABC are proportional to sides PQ, QR, and median PM of ΔPQR then ΔABC ∼ ΔPQR.

Let us prove that ΔABC ∼ ΔPQR with the given conditions.

**Explanation:**

**SAS property** states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Given that in ΔABC and ΔPQR,

AB / PQ = BC / QR = AD / PM

As AD and PM are median of ΔABC and ΔPQR respectively:

Therefore, BD / QM = [BC × 2] / [QR × 2] = BC / QR

Now, in ΔABD and ΔPQM

AB / PQ = BD / QM = AD / PM

Hence, ΔABD ∼ ΔPQM

Now in ΔABC and ΔPQR

AB / PQ = BC / QR [Given in the statement]

∠ABC = ∠PQR [ΔABD ∼ ΔPQM]

ΔABC ∼ ΔPQR [SAS Criterion]

### Hence proved.

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