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.
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]