Question

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR. 

Answer 1

Given that,

`("AB")/("PQ") = ("AC")/("PR") = ("AD")/("PM")`

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE          ...(By construction)

And, PM = ML          ...(By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC        ...(Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that

`("AB")/("PQ") = ("AC")/("PR") = ("AD")/("PM")`

`=>("AB")/("PQ")=("BE")/("QL")= (2"AD")/(2"PM")`

`=>("AB")/("PQ") = ("BE")/("QL") = ("AE")/("PL")`

∴ ΔABE ∼ ΔPQL        ...(By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL       ...(1)

Similarly, it can be proved that ΔAEC ∼ ΔPLR and

∠CAE = ∠RPL           ...(2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ              ...(3)

In ΔABC and ΔPQR,

`("AB")/("PQ")=("AC")/("PR")`

∠CAB = ∠RPQ               ...[Using equation (3)]

∴ ΔABC ∼ ΔPQR           ...(By SAS similarity criterion)