# 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 Δ ABC ~ Δ PQR

To Prove: Δ ABC ∼ Δ PQR
Given:

Proof:

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)

Diagonals AE and BC bisect each other at point D.

Therefore,

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)
Hence, Proved.

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