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{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} $>
(ii) $ \triangle \mathrm{DCB} \sim \triangle \mathrm{HGE} $>
(iii) $ \triangle \mathrm{DCA} \sim \triangle \mathrm{HGF} $

Given:

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. 

$∆ABC \sim ∆FEG$

To do:

We have to show that

(i) \( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \)

(ii) \( \triangle \mathrm{DCB} \sim \triangle \mathrm{HGE} \)

(iii) \( \triangle \mathrm{DCA} \sim \triangle \mathrm{HGF} \)

Solution:

(i)

$\triangle ABC$ and $\triangle FEG$

This implies,

$\angle A=\angle F$

$\angle B=\angle E$          

$\angle C=\angle G$

$\frac{AB}{FE}=\frac{BC}{EG}=\frac{AC}{FG}$

In $\triangle ACD$ and $\triangle FGH$,

$\angle A=\angle F$

$\angle 1=\angle 2$                    (Since $\frac{1}{2}\angle C=\frac{1}{2}\angle G$)

Therefore, by AA criterion,

$\triangle ACD \sim \triangle FGH$

This implies,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \)

Hence proved. 

(ii) $\triangle ACD \sim \triangle FGH$

This implies,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \)

\( \frac{\mathrm{AC}}{\mathrm{FG}}=\frac{\mathrm{BC}}{\mathrm{EG}} \)

 Therefore,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{BC}}{\mathrm{EG}} \)

In $\triangle BCD$ and $\triangle EGH$,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{BC}}{\mathrm{EG}} \)

$\angle 3=\angle 4$                    (Since $\frac{1}{2}\angle C=\frac{1}{2}\angle G$)

Therefore, by SAS criterion,

$\triangle DCB \sim \triangle HGE$

Hence proved.

(iii) $\triangle ACD \sim \triangle FGH$

This implies,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \)

\( \frac{\mathrm{AC}}{\mathrm{FG}}=\frac{\mathrm{BC}}{\mathrm{EG}} \)

 Therefore,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{BC}}{\mathrm{EG}} \)

In $\triangle DCA$ and $\triangle HGF$,

\( \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \)

$\angle 1=\angle 2$                    (Since $\frac{1}{2}\angle C=\frac{1}{2}\angle G$)

Therefore, by SAS criterion,

$\triangle DCA \sim \triangle HGF$

Hence proved.