Question

$\mathrm{CD}$ and $\mathrm{GH}$are respectively the bisectors of $\angle \mathrm{ACB}$ and $\angle \mathrm{EGF}$ such that $\mathrm{D}$ and $\mathrm{H}$ lie on sides$\mathrm{AB}$ and $\mathrm{FE}$ of $\mathrm{\Delta ABC}$ and $\mathrm{\Delta EFG}$

respectively. If $\mathrm{\Delta ABC}~\mathrm{\Delta FEG}$,

Show that:

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

(ii) $\mathrm{\Delta DCB}~\mathrm{\Delta HGE}$

(iii) $\mathrm{\Delta DCA}~\mathrm{\Delta HGF}$

Answer 1

Similarity of triangle:

$\mathrm{CD}$ and $\mathrm{GH}$ are respectively the bisectors of $\angle \mathrm{ACB}$ and $\angle \mathrm{EGF}$ such that $\mathrm{D}$ and $\mathrm{H}$ lie on sides $\mathrm{AB}$ and $\mathrm{FE}$ of $\mathrm{\Delta ABC}$ and $\mathrm{\Delta EFG}$ respectively.

To Prove

If $\mathrm{\Delta ABC}~\mathrm{\Delta FEG}$,

(i) $\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$
(ii)$\mathit{}\mathrm{\Delta DCB}~\mathrm{\Delta HGE}$
(iii) $\mathrm{\Delta DCA}~\mathrm{\Delta HGF}$

Proof

(i) We have,

$\mathrm{\Delta ABC}~\mathrm{\Delta FEG}$.

Therefore,

$\angle \mathrm{A}=\angle \mathrm{F},\angle \mathrm{B}=\angle \mathrm{E}$, and $\angle \mathrm{ACB}=\angle \mathrm{FGE}$

Now, since,

$\angle \mathrm{ACB}=\angle \mathrm{FGE}$

$\therefore \angle \mathrm{ACD}=\angle \mathrm{FGH}$ (Angle bisector)

And, $\angle \mathrm{DCB}=\angle \mathrm{HGE}$

In $\mathrm{\Delta ACD}$and $\mathrm{\Delta FGH}$,

$\begin{array}{rcl}\angle \mathrm{A}& =& \angle \mathrm{F}\\ \angle \mathrm{ACD}& =& \angle \mathrm{FGH}\end{array}$

By $\mathrm{AA}$similarity criterion, we have,

$$\begin{gathered} \mathrm{\Delta ACD}~\mathrm{\Delta FGH} \\ \therefore \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}} \end{gathered}$$

(ii) In $\mathrm{\Delta DCB}$ and $\mathrm{\Delta HGE}$,

Proved in part (i)

$\begin{array}{rcl}\angle \mathrm{DCB}& =& \angle \mathrm{HGE}\\ \angle \mathrm{B}& =& \angle \mathrm{E}\end{array}$

By$\mathrm{AA}$ similarity criterion,

$\therefore \mathrm{\Delta DCB}~\mathrm{\Delta HGE}$

(iii) In $\mathrm{\Delta DCA}$ and $\mathrm{\Delta HGF}$,

Since, we have already proved,

$\begin{array}{rcl}\angle \mathrm{ACD}& =& \angle \mathrm{FGH}\\ \angle \mathrm{A}& =& \angle \mathrm{F}\end{array}$

By $\mathrm{AA}$similarity criterion,

$\therefore \mathrm{\Delta DCA}~\mathrm{\Delta HGF}$

Hence Proved