# 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 $âˆ†ABC$ and $âˆ†EFG$ respectively. If $âˆ†ABC \sim âˆ†FEG$, show that (i) $\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$>(ii) $\triangle \mathrm{DCB} \sim \triangle \mathrm{HGE}$>(iii) $\triangle \mathrm{DCA} \sim \triangle \mathrm{HGF}$

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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.

Updated on 10-Oct-2022 13:21:14