CD and GH are respectively the bisectorsof ∠ ACB and ∠ EGF in such a way that D and H lieon sides AB and FE of Δ ABC and Δ EFGrespectively. If Δ ABC ~ Δ FEG, show that:

(i)

(ii) Δ DCB ~ Δ HGE

(iii) Δ DCA ~ Δ HGF

It is given that ΔABC ΔFEG

∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

(i) In ΔACD and ΔFGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ΔACD ΔFGH (By AA similarity criterion)

(Corresponding sides of similar triangles are proportional)

(ii) In triangle DCB and HGE

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

Therefore,

ΔDCB ΔHGE (By AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

ΔDCA ΔHGF (By AA similarity)