Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $∆PQR$ (see in given figure). Show that $∆ABC \sim ∆PQR$.

Given:

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $∆PQR$.

To do:

We have to show that $∆ABC \sim ∆PQR$.

Solution:

In $\triangle ABC$ and $\triangle PQR$,

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

Therefore, by SSS criterion,

$\triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}$

This implies,

$\angle B=\angle Q$             (CPCT)

In $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$,

$\angle B=\angle Q$

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}$

Therefore, by SAS criterion,

$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$