137
CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\triangle ABC$ and $\triangle EFG$ respectively.
If $\triangle ABC \sim \triangle FEG$, then show that
(i) $\frac{CD}{GH} = \frac{AC}{FG}$ and
(ii) $\triangle DCB \sim \triangle HGE$.
If $\triangle ABC \sim \triangle FEG$, then show that
(i) $\frac{CD}{GH} = \frac{AC}{FG}$ and
(ii) $\triangle DCB \sim \triangle HGE$.
Show SolutionHide Solution↓
(i) $\triangle ABC \sim \triangle FEG$ (given)
$\angle ACB = \angle FGE$
$\frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE \Rightarrow \angle ACD = \angle FGH$ ($\frac{1}{2}$ Mark)
Also, $\angle A = \angle F$
$\therefore \triangle ACD \sim \triangle FGH$ (2 Marks)
$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$ ($\frac{1}{2}$ Mark)
(ii) In $\triangle DCB$ and $\triangle HGE$
$\angle DBC = \angle HEG$ (1 Mark)
$\angle DCB = \angle HGE$ (1 Mark)
$\therefore \triangle DCB \sim \triangle HGE$
$\angle ACB = \angle FGE$
$\frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE \Rightarrow \angle ACD = \angle FGH$ ($\frac{1}{2}$ Mark)
Also, $\angle A = \angle F$
$\therefore \triangle ACD \sim \triangle FGH$ (2 Marks)
$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$ ($\frac{1}{2}$ Mark)
(ii) In $\triangle DCB$ and $\triangle HGE$
$\angle DBC = \angle HEG$ (1 Mark)
$\angle DCB = \angle HGE$ (1 Mark)
$\therefore \triangle DCB \sim \triangle HGE$