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 $\Delta ABC\sim \Delta FEG$, prove that
$\begin{align}
  & (i)\text{ }\Delta DCA\sim \Delta HGF \\
 & \\
 & (ii)\text{ }\dfrac{CD}{GH}=\dfrac{AC}{FG} \\
 & \\
 & (iii)\text{ }\Delta DCB\sim \Delta HGE \\
\end{align}$

Answer 1

Hint:Angle bisector divides the angle into 2 equal halves: consider $\Delta DCA$ and $\Delta HGF$ , BY $AA$ Similarly prove they are similar. By taking corresponding sides of a similar triangle, prove $\dfrac{CD}{GH}=\dfrac{AC}{FG}$ . Now prove $\Delta DCB$ and $\Delta HGE$ are similar by $AA$ similarity criterion.

Complete step-by-step answer:
We have been given two triangles, which are $\Delta ACB$ and $\Delta EGF$ respectively. It is given that CD is the bisector of $\angle ACB$ . An angle bisector is a line that divides an angle into two congruent angles. As CD bisects $\angle ACB$ , we can say that $\angle ACD=\angle BCD$ , from the figure.
Thus we can say that, $\angle ACD=\angle BCD=\dfrac{1}{2}\angle ACB.$
Similarly we can say that GH is the bisector of $\angle EGF$ . Hence GH divides the angle into two congruent angles. Thus we can say that,
$\angle FGH=\angle EGH=\dfrac{1}{2}\angle EGF$ .
We have been also told that $\Delta ABC\sim \Delta FEG.$
$(i)$ Prove that, $\Delta DCA\sim \Delta FEG$ .
We said that $\Delta ABC$ and $\Delta FEG$ are similar. The corresponding angles of similar triangles are equal. Thus we can say that,
 $\angle A=\angle F$ (Corresponding angles).
$\angle C=\angle G$ (Corresponding angles).

We said that CD bisects $\angle C$ and GH bisects $\angle G$
GH bisects $\angle G$ .
There we can say that,
$\angle C=\angle G\Rightarrow \dfrac{1}{2}\angle C=\dfrac{1}{2}\angle G$ .
We said that, $\angle ACD=\angle BCD=\dfrac{1}{2}\angle C.$ and $\angle FGH=\angle EGH=\dfrac{1}{2}\angle G.$
Thus, $\dfrac{1}{2}\angle C=\dfrac{1}{2}\angle G$ .
$\Rightarrow \angle ACD=\angle FGH.$
Since CD and GH are bisector of $\angle C$ and $\angle G$ we can say that $\angle ACD=\angle FGH.$
Now, let us consider $\Delta ACD$ and $\Delta FGH.$
$\angle A=\angle F$ (Corresponding angle).
$\angle ACD=\angle FGH$ (Proved above).
Now, two angles of $\Delta ACD$ is equal to two angles of $\Delta FGH.$ This by AA similarity criterion, we can say that both the triangles are similar.
i.e. $\Delta ACD\sim \Delta FGH.$
Hence we proved that $\Delta ACD$ is similar to $\Delta FGH.$
$(ii)$ To prove \[\dfrac{\text{CD}}{GH}=\dfrac{\text{AC}}{FG}\] .
Let us consider $\Delta ACB$ and $\Delta EFG$ . We proved earlier that, $\angle C=\angle G\Rightarrow \dfrac{1}{2}\angle C=\dfrac{1}{2}\angle G$ .
$\therefore \angle ACD=\angle FGH$ , as CD and GH are bisectors of $\angle C$ and $\angle G$ .
Now let us consider $\Delta ACD$ and $\Delta FGH$ , by AA similarly, there triangle are similar
$\Delta ACD\sim \Delta FGH$ .
We know that corresponding sides of similar triangles are proportional.
There from $\Delta ACD$ and $\Delta FGH$ , as they are similar, we can say that $\dfrac{\text{CD}}{\text{GH}}=\dfrac{\text{AC}}{GF}$ .
Thus we proved that they are in proportion.

$(iii)$ To prove that $\Delta DCB\sim \Delta HGE$ we proved that $\Delta ACB\sim \Delta EFG$ .
Thus the corresponding angles of the similar triangle are equal.
Thus we can say that $\angle B=\angle E$ (from figure).
We said that CD bisects $\angle C$ and $GH$ bisects $\angle G$ .
Thus we say that, $\angle C=\angle G=\dfrac{1}{2}\angle C=\dfrac{1}{2}\angle G$ .
 \[\begin{align}
  & i.e.\text{ }\angle ACD=\angle BCD=\dfrac{1}{2}\angle C. \\
 & \angle FGH=\angle EGH=\dfrac{1}{2}\angle G. \\
\end{align}\]
Thus ,$\dfrac{1}{2}\angle C=\dfrac{1}{2}\angle G\Rightarrow \angle BCD=\angle EGH.$
Now let us consider $\Delta DCB$ and $\Delta HGE$ .
$\angle B=\angle E$ (corresponding angle)
$\angle BCD=\angle EGH$ (proved above).
Now two angles of $\Delta DCB$ is equal to two angles of $\Delta AGE.$
Thus by AA similarity criterion, we can say that both triangles are similar.
$i.e.\Delta DCB\sim \Delta HGE$ .
Thus we proved that $\Delta DCB$ is similar to $\Delta HGE$ .

Note: As we are already given that $\Delta ACB$ and $\Delta FGE$ are similar, we can make out that $\Delta DAC$ and $\Delta FGH$ are similar and $\Delta DCB$ & and $\Delta HGE$ are similar with the help of similarly criterion we can prove the same.