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 ∆ EFG respectively. If ∆ ABC ~ ∆ FEG, show that:
(i) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) ∆ DCB ~ ∆ HGE
(iii) ∆ DCA ~ ∆ HGF

(i) This is referred to as the AA similarity criterion for two triangles.

∠ACB = ∠FGE (Given that ∆ABC ~ ∆FEG)

⇒ ∠ACB/2 = ∠FGE/2

⇒ ∠ACD = ∠FGH (CD and GH are bisectors of ∠C and ∠G respectively) ------------ (1)

In ∆ADC and ∆FHG

∠DAC = ∠HFG [∆ADC ~ ∆FEG]

∠ACD = ∠FGH [From equation (1)]

Thus, ∆ADC ~ ∆FHG (AA criterion)

⇒$\frac{CD}{GH} = \frac{AC}{FG}$  [If two triangles are similar, then their corresponding sides are in the same ratio.]

(ii) In ∆DCB and ∆HGE

∠DBC = ∠HEG [∆ABC ~ ∆FEG]

∠DCB = ∠HGE [∵ ∠ACB/2 = ∠FGE/2]

Thus, ∆DCB ~ ∆HGE (AA criterion)

(iii) In ∆DCA and ∆HGF

∠DAC = ∠HFG [∆ABC ~ ∆FEG]

∠ACD = ∠FGH [∵ ∠ACB = ∠FGE]

Thus, ∆DCA ~ ∆HGF (AA criterion)