Question

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that `("AB")/("PQ") = ("AD")/("PM")`.

Answer 1

Given, ΔABC ∼ ΔPQR

⇒ `("AB")/("PQ") = ("BC")/("QR") = ("AC")/("PR")`

(From the side-ratio property of similar triangles)

⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R       ...(i)

BC = 2BD and QR = 2QM         ...(∵ P and M are the midpoints of BC and QR)

⇒ `("AB")/("PQ") = (2  "BD")/(2  "QM") = ("AC")/("PR")`

⇒ `("AB")/("PQ") = ("BD")/("QM") = ("AC")/("PR")`      ...(ii)

Now in ΔABD and ΔPQM

`("AB")/("PQ") = ("BD")/("QM")`        ...(From (ii))

∠B = ∠Q       ...(From (i))

⇒ ΔABD ∼ ΔPQM         ...(By SAS similarity criterion)

∴ `("AB")/("PQ") = ("AD")/("PM")`

Hence, proved.