Sides $\mathrm{AB}$ and $\mathrm{BC}$ and median $\mathrm{AD}$ of a triangle $\mathrm{ABC}$ are respectively proportional to sides $\mathrm{PQ}$ and $\mathrm{QR}$ and median $\mathrm{PM}$ of $\triangle \mathrm{PQR}$ (see figure).
Question.

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

Solution:

As, $\frac{A B}{P O}=\frac{B C}{O R}=\frac{A D}{P M}$

So, $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{OM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$

$\left\{\because \frac{A B}{P Q}=\frac{\frac{1}{2} B C}{\frac{1}{2} O R}=\frac{B D}{O M}\right\}$

$\therefore$ By SSS similarity,

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

$\mathrm{As}, \Delta \mathrm{ABD} \sim \triangle \mathrm{PQM}$

$\therefore \quad \angle \mathrm{ABD}=\angle \mathrm{PQM}$

Now, In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$

$\frac{A B}{P Q}=\frac{B C}{Q R}$ (Given)

$\angle \mathrm{ABC}=\angle \mathrm{PQR}($ Proved above $)$

$\therefore$ By SAS similarity

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