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)


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