# 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 $$\triangle$$ ABC and $$\triangle$$ EFG respectively. If $$\triangle$$ ABC ~ $$\triangle$$ FEG show that (i) $$CD\over GH$$ = $$AC\over FG$$ (ii) $$\triangle$$ DCB ~ $$\triangle$$ HGE (iii) $$\triangle$$ DCA ~ $$\triangle$$ HGF

## Solution :

Given : $$\triangle$$ ABC ~ $$\triangle$$  FEG and CD and GH are bisectors of $$\angle$$ C and $$\angle$$ G respectively.

(i) In triangle ACD and FGH

$$\angle$$ A = $$\angle$$ F         ($$\triangle$$ ABC ~ $$\triangle$$  FEG)

Since it is given that CD bisects $$\angle$$ C and GH bisects $$\angle$$ G and $$\angle$$ C = $$\angle$$ G,

$$\implies$$   $$\angle$$ ACD = $$\angle$$ FGH

$$\therefore$$  By AA similarity,

$$\triangle$$ ACD ~ $$\triangle$$ FGH

$$\implies$$ $$CD\over GH$$ = $$AC\over FG$$

(ii)  In triangle DCB and HGE,

$$\angle$$ B = $$\angle$$ E         ($$\triangle$$ ABC ~ $$\triangle$$  FGH)

Since it is given that CD bisects $$\angle$$ C and GH bisects $$\angle$$ G and $$\angle$$ C = $$\angle$$ G,

$$\implies$$   $$\angle$$ DCB = $$\angle$$ HGE

$$\therefore$$  By AA similarity,

$$\triangle$$ DCB ~ $$\triangle$$ HGE

(iii) In triangle DCA and HGF

$$\angle$$ A = $$\angle$$ F         ($$\triangle$$ ABC ~ $$\triangle$$  FEG)

Since it is given that CD bisects $$\angle$$ C and GH bisects $$\angle$$ G and $$\angle$$ C = $$\angle$$ G,

$$\implies$$   $$\angle$$ DCA = $$\angle$$ HGF

$$\therefore$$  By AA similarity,

$$\triangle$$ DCA ~ $$\triangle$$ HGF