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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 is similar to triangle FEG, show that: (question 1 )

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:

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

Views

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

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