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 : triangles\(\triangle\) 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

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