To Prove: Δ ABC ∼ Δ PQR
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.
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,
∠CAB = ∠RPQ [Using equation (iii)]
ΔABC ΔPQR (By SAS similarity criterion)
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