# If AD and PM are medians of triangles ABC and PQR respectively, where $$\triangle$$ ABC ~ $$\triangle$$ PQR, prove that $$AB\over PQ$$ = $$AD\over PM$$.

## Solution :

Given : AD and PM are the medians of triangles ABC and PQR respectively, where $$\triangle$$ ABC ~ $$\triangle$$ PQR.

To Prove : $$AB\over PQ$$ = $$AD\over PM$$

Proof  : In triangles ABD and PQM, we have

$$\angle$$ B = $$\angle$$ D        (because $$\triangle$$ ABC ~ $$\triangle$$ PQR)

Since AD and PM are the medians to BC and QR respectively and $$AB\over PQ$$ = $$BC\over QR$$

$$AB\over PQ$$ = $${1\over2}BC\over {1\over 2}QR$$

So, $$AB\over PQ$$ = $$BD\over QM$$

$$\therefore$$  By SAS similarity,

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

So, $$AB\over PQ$$ = $$BD\over QM$$ = $$AD\over PM$$