Question

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR show that ΔABC ~ ΔPQR.

Answer 1

We have MBC and PQR in which AD and PM are medians corresponding to sides BC and QR respectively such that

`("AB")/("PQ") = ("BC")/("QR") = ("AD")/("PM")`

`("AB")/("PQ") = (1/2 "BC")/(1/2 "QR") = ("AD")/("PM")`

`("AB")/("PQ") = ("BD")/("QM") = ("AD")/("PM")`

Using SSS similarity, we have

ΔABD ~ ΔPQM

Their corresponding angles are equal.

∠ABD = ∠PQM

∠ABC = ∠PQR

Now, in MBC and ΔPQR,

`("AB")/("PQ") = ("BC")/("QR")`          ...[1]

Also, ∠ABC = ∠PQR              ...[2]

From [1] and [2]

ΔABC ~ ΔPQR         ...[SAS similarity]