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# 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/GH = AC/FG (ii) $$\triangle$$ DCB ~ $$\triangle$$ HGE (iii) $$\triangle$$ DCA ~ $$\triangle$$ HGF

Given, 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.

(i) From the given condition,

$$\triangle$$ ABC ~ $$\triangle$$ FEG.
$$\angle$$ A = $$\angle$$ F, $$\angle$$ B = $$\angle$$ E, and $$\angle$$ ACB = $$\angle$$ FGE

Since, $$\angle$$ ACB = $$\angle$$ FGE
$$\angle$$ ACD = $$\angle$$ FGH (Angle bisector)
And, $$\angle$$ DCB = $$\angle$$ HGE (Angle bisector)

In $$\triangle$$ ACD and $$\triangle$$ FGH,
$$\angle$$ A = $$\angle$$ F
$$\angle$$ ACD = $$\angle$$ FGH
$$\triangle$$ ACD ~ $$\triangle$$ FGH (AA similarity criterion)
CD/GH = AC/FG

(ii) In $$\triangle$$ DCB and $$\triangle$$ HGE,
$$\angle$$ DCB = $$\angle$$ HGE (Already proved)
$$\angle$$ B = $$\angle$$ E (Already proved)
$$\triangle$$ DCB ~ $$\triangle$$ HGE (AA similarity criterion)

(iii) In $$\triangle$$ DCA and $$\triangle$$ HGF,
$$\angle$$ ACD = $$\angle$$ FGH (Already proved)
$$\angle$$ A = $$\angle$$ F (Already proved)
$$\triangle$$ DCA ~ $$\triangle$$ HGF (AA similarity criterion)