Sides AB and AC and median AD of atriangle ABC are respectivelyproportional to sides PQ and PR andmedian PM of another triangle PQR.

Given that,

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.

Hence, 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 in the question that

ΔABE  ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal

∠BAE = ∠QPL     (i)

Similarly, it can be proved that

ΔAEC ~ ΔPLR and

∠CAE = ∠RPL     (ii)

Adding equation (i) and (ii), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒∠CAB = ∠RPQ     (iii)

In ΔABC and ΔPQR,

     (Given)

∠CAB = ∠RPQ [Using equation (iii)]

ΔABC ~ ΔPQR (By SAS similarity criterion)