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)