$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\Delta ABC$ and $\Delta EFG$ respectively. If $\Delta ABC \sim \Delta FEG,$ show that:
$(i) \frac{CD}{GH} = \frac{AC}{FG}$
$(ii) \Delta DCB \sim \Delta HGE$
$(iii) \Delta DCA \sim \Delta HGF$

(N/A) It is given that $\Delta ABC \sim \Delta FEG$.
Therefore,$\angle A = \angle F, \angle B = \angle E,$ and $\angle ACB = \angle FGE$.
Since $\angle ACB = \angle FGE,$ their bisectors are also equal.
Therefore,$\angle ACD = \angle FGH$ (Angle bisector).
And,$\angle DCB = \angle HGE$ (Angle bisector).
$(i)$ In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (Given)
$\angle ACD = \angle FGH$ (Proved above)
Therefore,$\Delta DCA \sim \Delta HGF$ (By $AA$ similarity criterion).
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{CD}{GH} = \frac{AC}{FG}$.
$(ii)$ In $\Delta DCB$ and $\Delta HGE$:
$\angle DCB = \angle HGE$ (Proved above)
$\angle B = \angle E$ (Given)
Therefore,$\Delta DCB \sim \Delta HGE$ (By $AA$ similarity criterion).
$(iii)$ In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (Given)
$\angle ACD = \angle FGH$ (Proved above)
Therefore,$\Delta DCA \sim \Delta HGF$ (By $AA$ similarity criterion).