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Q10 (1)   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 \Delta ABC\: \: and\: \: \Delta EGF respectively. If \Delta ABC \sim \Delta EGF, show that:

              \frac{CD}{GH} = \frac{AC}{FG} 

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best_answer

To prove : 

                 \frac{CD}{GH} = \frac{AC}{FG}

Given : \Delta ABC \sim \Delta EGF

\angle A=\angle F,\angle B=\angle E\, \, and \, \, \angle ACB=\angle FGE,\angle ACB=\angle FGE

\therefore \angle ACD=\angle FGH    ( CD and GH are bisectors of equal angles)

\therefore \angle DCB=\angle HGE   ( CD and GH are bisectors of equal angles)

In \Delta ACD \, \, and\, \, \Delta FGH

\therefore \angle ACD=\angle FGH     ( proved above)

    \angle A=\angle F                        ( proved above)

\Delta ACD \sim \Delta FGH       ( By AA criterion)

\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}

Hence proved 

 

 

 

 

Posted by

seema garhwal

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