Question

$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $△ABC$ and $△EFG$ respectively. If $△ABC\sim △FEG$, show that:
(i) $\frac{CD}{GH}=\frac{AC}{FG}$
(ii) $△DCB\sim △HGE$
(iii) $△DCA\sim △HGF$

Answer 1

In $△ABC$ and $△FEG,$
`$△ABC\sim FEG$
$\therefore$ $\angle ACB=\angle EGF$ (Corresponding angles of similar triangles)
Since, $DC$ and $GH$ are bisectors of $\angle ACB$ and $\angle EGH$ respectively.
$\therefore$ $\angle ACB=2\angle ACD=2\angle BCD$
And $\angle EGF=2\angle FGH=2\angle HGE$
$\therefore$ $\angle ACD=\angle FGH$ and $\angle DCB=\angle HGE$ $...................\left(1\right)$
Also $\angle A=\angle F$ and $\angle B=\angle E$ $...............\left(2\right)$
In $△ACD$ and $△FGH,$
$\angle A=\angle F$ (From $2$)
$\angle ACD=\angle FGH$ (From $1$)
$\therefore$ By AA criterion of similarity $△ACD$ $\sim$$△FGH$

$△DCA$ $\sim$$△HGF$ [(i) and (iii) proved]

$\therefore$ $\frac{CD}{GH}=\frac{AC}{FG}$ (Corresponding Sides of Similar Triangles)
In $△DCB$ and $△HGE,$
$\angle B=\angle E$ (From 2)
$\angle DCB=\angle HGE$ (From 1)
$\therefore$ By AA criterion of similarity $△DCB$ $\sim$$△HGE$ [(ii) proved]