Question

Question 16
If AD and PM are medians of triangles ABC and PQR, respectively where $\Delta ABC\sim \Delta PQR$. Prove that $\frac{AB}{PQ}=\frac{AD}{PM}$.

Answer 1

It is given that $\Delta ABC\sim \Delta PQR$
We know that the corresponding sides of similar triangles are proportional.
$\therefore \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}...\left(i\right)$
Also, $\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R...\left(ii\right)$
Since AD and PM are medians, they will divide their opposite sides.
$\therefore BD=\frac{BC}{2}andQM=\frac{QR}{2}....\left(iii\right)$
From equations (i) and (iii), we get
$\frac{AB}{PQ}=\frac{BD}{QM}...\left(iv\right)$
In $\Delta ABDand\Delta PQM$,
$\angle B=\angle Q$ [Using equation (ii)]
$\frac{AB}{PQ}=\frac{BD}{QM}$ [Using equation (iv)]
$\therefore \Delta ABD\sim \Delta PQM$ (By SAS similarity criterion)
$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$.