Question

In given figures if CD and GH (D and H lie on AB and FE) are respectively bisectors of $\angle ACB$ and $\angle EGF$ and $△ABC\sim △FEG$, prove that

(i) $△DCA\sim △HGF$

(ii) $\frac{CD}{GH}=\frac{AC}{FG}$

(iii) $△DCB\sim △HGE$

Answer 1

(i) We have ,

$△ABC\sim △FEG$

$\Rightarrow$ $\angle A=\angle F$

and, $\angle C=\angle G$........(i)

$\Rightarrow$ $\frac{1}{2}\angle C=\frac{1}{2}\angle G$

$\Rightarrow$ $\angle 1=\angle 3$ and $\angle 2=\angle 4$ [$\therefore$ CD and GH are bisector of $\angle C$ and $\angle G$ respectively]......(ii)

Thus, in ${△}^{\prime }sACD$ and $FGH$, we have

$\angle A=\angle F$ [From (i)]

$\angle 2=\angle 4$ [From (ii)]

Therefore, by AA-criterion of similarity, we have

$△ACD\sim △FGH$ or, $△DCA\sim △HGF$ $\left[Henceproved\right]$

(ii) We have,

$△ACD\sim △FGH$

$\Rightarrow$ $\frac{AC}{FG}=\frac{CD}{GH}$ $\left[Henceproved\right]$

(iii) In ${△}^{\prime }sDCB$ and $HGE$, we have

$\angle 1=\angle 3$ [From (ii)]

$\angle B=\angle E$ [$\because △ABC\sim △FEG$]

Thus, by AA-criterion of similarity, we have

$△DCB\sim △HGE$ $\left[Henceproved\right]$