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