Question

In the adjoining figure, prove that $△FGH$ and $△QPR$ are similar.

Answer 1

In $△$ FGH,

$\angle F={180}^{\circ }-(\angle G+\angle H)\angle F={180}^{\circ }-({20}^{\circ }+{30}^{\circ })\angle F={130}^{\circ }$

$In\Delta FGHand\Delta QPR,$
$\angle F=\angle Q={130}^{\circ }$
$\frac{FG}{PQ}=\frac{63}{7}=\frac{9}{1}$
$\frac{FH}{QR}=\frac{45}{5}=\frac{9}{1}$

$\Rightarrow \frac{FG}{PQ}=\frac{FH}{QR}$

Hence, $△FGH\sim △QPR$ by SAS similarity.