Given : In two triangles ABC and ∆PQR D is mid-point of BC and M is mid-point of QR.

and \(\frac { AB }{ PQ }\) = \(\frac { AC }{ PR }\) = \(\frac { AD }{ PM }\) ……(i)

To prove : ∆ABC ~ ∆PQR

Construction : Produce AD up to E such that AD = DE. Join BE and CE and produce PM upto N such that PM = MN. Join QN and NR.

Proof : Diagonals AE and BC of quadrilateral ABEC bisect each other at point D.

∴ quadrilateral ABEC is a parallelogram.

∴ BE = AC …(ii)

Similarly PQNR is a parallelogram

∴ QN = PR …..(iii)

Dividing equation (ii) by (iii)

\(\frac { BE }{ QN }\) = \(\frac { AC }{ PR }\) …..(iv)

Now, \(\frac { AD }{ PM }\) = \(\frac { 2AD }{ 2PM }\) = \(\frac { AD }{ PM }\) = \(\frac { AE }{ PN }\) …(v)

In equation (i), (iv) and (v),

\(\frac { AB }{ PQ }\) = \(\frac { BE }{ QN }\) = \(\frac { AE }{ PN }\)

Thus, in ∆ABE and ∆PQN

\(\frac { AB }{ PQ }\) = \(\frac { BE }{ QN }\) = \(\frac { AE }{ PN }\)

∴ ∆ABE ~ ∆PQN (By SSS)

∴ ∠BAE = ∠QPN …(vi)

Similarly

∆AEC ~ ∆PNR

∠EAC = ∠NPR …(vii)

Adding equation (vi) and (vii)

∠BAE + ∠EAC = ∠QPN + ∠NPR

⇒ ∠BAC = ∠QPR

Now, in ∆ABC and ∠PQR

\(\frac { AB }{ PQ }\) = \(\frac { AC }{ PR }\) [From equation (i)]

∠A = ∠P

∴ By SAS similarity criterion

∆ABC ~ ∆PQR