CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆FEG respectively. If ∆ABC ~ ∆FEG then

show that
(i) \frac{CD}{GH} = \frac{AC}{FG}
(ii) ∆DCB ~ ∆HGE
show that
(i) \frac{CD}{GH} = \frac{AC}{FG}
(ii) ∆DCB ~ ∆HGE
Step 1: Understand the question and According to the given information make a imaginary diagram.
Step 2: Note down the equal angles in both the triangles (ΔABC ~ ΔFEG)
NOTE: It is given that ΔABC ~ ΔFEG.
So, \angle A = \angle F ,
\angle B = \angle E
\angle B = \angle G
Step 3: Find the similar triangles hidden in the figure. (ΔACD ~ ΔFGH).
NOTE: We know that \angle ACD = \angle FGH (Angle bisector)
From this We can write \angle DCB = \angle HGE (Angle bisector)
\angle A = \angle F (Proved above)
∴ ΔACD ~ ΔFGH (By AA similarity criterion)
Step 4: Prove the required ratio
NOTE: ΔACD ~ ΔFGH
So, their corresponding sides are in the same ratio.
\frac{CD}{GH} = \frac{AC}{FG}
Condition (i) proved.
Step 5: Find the similar figures hidden in the figure (ΔDCB ~ ΔHGE)
NOTE: In ΔDCB and ΔHGE,
\angle DCB = \angle HGE (Proved above)
\angle B = \angle E (Proved above)
∴ ΔDCB ~ ΔHGE (By AA similarity criterion)
Condition (ii) proved