# If AD and PM are medians of triangles ABC and PQR respectively, where$âˆ†ABC \sim âˆ†PQR$. Prove that $\frac{AB}{PQ}=\frac{AD}{PM}$âˆ™

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Given:

$AD$ and $PM$ are medians of $\vartriangle ABC$ and $\vartriangle PQR$ respectively where, $\vartriangle ABC\sim \vartriangle PQR$.

To do:

We have to prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.

Solution:

It is given that $\vartriangle ABC \sim \vartriangle PQR$

We know that,

The corresponding sides of similar triangles are always proportional.

This implies,

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$............(i)

$\angle A = \angle P,\ \angle B = \angle Q,\ \angle C = \angle R$............(ii)

$AD$ and $PM$ are medians.

This implies,

They divide their opposite sides $BC$ and $QR$ respectively.

Therefore,

$BD=\frac{BC}{2}$ $QM=\frac{QR}{2}$.......(iii)

From equations (i) and (iii), we get,

$\frac{AB}{PQ}=\frac{BD}{QM}$.............(iv)

In $\vartriangle ABD$ and $\vartriangle PQM$,

$\angle B = \angle Q$         [From (ii)]

$\frac{AB}{PQ}=\frac{BD}{QM}$     [From (iv)]

Therefore, by SAS similarity,

$\vartriangle ABD \sim \vartriangle PQM$

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

Hence proved.

Updated on 10-Oct-2022 13:21:19