# Ex.6.3 Q10 Triangles Solution - NCERT Maths Class 10

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## Question

$$CD$$ and $$GH$$ are respectively the bisectors of $$\angle ACB$$ and $$\angle EGF$$ such that $$D$$ and $$H$$ lie on sides $$AB$$ and $$FE$$ of $$\Delta ABC$$ and $$\Delta EFG$$ respectively. If $$\triangle ABC \sim \triangle FEG$$, show that:

(i)\begin{align}\frac{CD}{GH}=\frac{AC}{FG}\end{align}

(ii)\begin{align}\text{ }\Delta DCB\text{ }\sim{\ }\text{ }\Delta HGE\end{align}

(iii) \begin{align}\Delta DCA\text{ }\sim{\ }\Delta HGF\end{align}

Diagram ## Text Solution

(i) Reasoning:

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This may be referred to as the $$AA$$ similarity criterion for two triangles.

Steps:

$$\angle ACB\,=\,\angle FGE$$

$$\Rightarrow \frac {{\angle ACB}}{2}= \frac {{\angle FGE}}{2}$$

$$\Rightarrow \angle ACD\,=\,\angle FGH$$    ($$CD$$ and GH are bisectors of $$\angle C {\,\rm{and}} \,\angle G$$ respectively)

In $$\Delta ADC\,{\rm{and}}\,\Delta FHG$$

\begin{align}\angle D A C&=\angle H F G\,\,[\because \Delta A D C \sim \Delta F E G]\\ \end{align}

$$\angle A C D=\angle F G H$$

\begin{align} \Rightarrow \Delta A D C\,\sim\,\Delta F H G \end{align}   ($$AA$$ criterion)

[If two triangles are similar, then their corresponding sides are in the same ratio]

\begin{align} \Rightarrow \frac{C D}{G H}=\frac{A C}{F G}\end{align}

(ii) Reasoning:

If two angles of one triangle are respectively equal to two angles of another triangle,then the two triangles are similar.

This is reffered as AA criterion for two triangles.

Steps:

In $$\Delta DCB$$ and $$\Delta HGE$$

\begin{align}\angle DBC&=\angle HEG\,\left(\because \Delta ABC\sim \Delta FEG \right) \\ \angle DCB&=\angle HGE\,\left(\because \frac{\angle ACB}{2}=\frac{\angle FGE}{2} \right) \\ \Rightarrow\qquad \Delta DCB &\sim \Delta EHG(AA\,\text{criterion}) \\ \end{align}

(iii) Reasoning:

If two angles of one triangle are respectively equal to two angles of another triangle,then the two triangles are similar.

This is reffered as AA criterion for two triangles.

Steps:

In $$\Delta DCA,\,\,\Delta HGF$$

\begin{align}& \angle DAC=\angle HFG\,\,\left[ \because\Delta ABC\sim \Delta FEG \right] \\ & \angle ACD=\angle FGH\,\,[\because \frac{{\angle ACB}}{2} = \frac {{\angle FGE}}{2}] \\ \Rightarrow\qquad &\Delta DCA\sim \Delta HGF\,\,({AA}\,\text{criterion}) \\ \end{align}

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