Question

Question 10 (i)
CD and GH are respectively the bisectors of $\angle ACBand\angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC$ and $\Delta EFG$ respectively. If $\Delta ABC\sim \Delta FEG$, Show that
$\left(i\right)\frac{CD}{GH}=\frac{AC}{FG}$

Answer 1

(i) It is given that $\Delta ABC\sim \Delta FEG$.
$\therefore \angle A=\angle F,\angle B=\angle E,and\angle ACB=\angle FGE$
$\therefore \angle ACD=\angle FGH$ (Angle bisector)
And, $\angle DCB=\angle HGE$ (Angle bisector)

In $\Delta ACDand\Delta FGH,$
$\angle A=\angle F$
$\angle ACD=\angle FGH$
$\therefore \Delta ACD\sim \Delta FGH$ (By AAA similarity criterion)
$\Rightarrow \frac{CD}{GH}=\frac{AC}{FG}$