# 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 $$\triangle$$ ABC ~ $$\triangle$$ PQR.

## Solution :

Given : $$\triangle$$ ABC and $$\triangle$$ PQR in which AD and PM are the medians, such that

$$AB\over PQ$$ = $$BC\over QR$$ = $$AD\over PM$$

To prove :  $$\triangle$$ ABC ~ $$\triangle$$ PQR

Proof : $$AB\over PQ$$ = $$BC\over QR$$ = $$AD\over PM$$

$$\implies$$  $$AB\over PQ$$ = $${1\over 2}BC\over {1\over 2}QR$$ = $$AD\over PM$$

$$\implies$$ $$AB\over PQ$$ = $$BD\over QM$$ = $$AD\over PM$$

By SSS similarity,

$$\triangle$$ ABD ~ $$\triangle$$ PQM

$$\implies$$ $$\angle$$ B = $$\angle$$ Q

Now, in triangle ABC and PQR, we have

$$AB\over PQ$$ = $$BC\over QR$$         (given)

and  $$\angle$$ B = $$\angle$$ Q

So, By SAS similarity,

$$\triangle$$ ABC ~ $$\triangle$$ PQR