Question

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

Answer 1

Consider the triangles $△ABC$ and $△PQR$

$AD$ and $PM$ being the mediums from vertex $A$ and $P$ respectively.

Given : $△ABC\sim △PQR$

To prove : $\frac{AB}{PQ}=\frac{AD}{PM}$

It is given that $△ABC\sim △PQR$

$\Rightarrow \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

[ from the side-ratio property of similar $△$ s]

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

$BC=2BD;QR=2QM$ $[P,M$ being the mid points of $BC$ $q$ $QR$ respectively]

$\Rightarrow \frac{AB}{PQ}=\frac{2BD}{2QM}=\frac{AC}{PR}$

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AC}{PR}$........(1)

Now in $△ABDq△PQM$

$\frac{AB}{PQ}=\frac{BP}{QM}$........[ from (1)]

$\angle B=\angle Q$........[ from (A)]

$\Rightarrow △ABD\sim △PQM$ [ By $SAS$ property of similar $△$ s] from the side property of similar $△$ s Hence proved

$\frac{AB}{PQ}=\frac{AD}{PM}$