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Answer :

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)

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