# Ex. 6.6 Q9 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.6'

## Question

In Fig.below, $$D$$ is a point on side $$BC$$ of $$\Delta ABC$$ such that \begin{align}\frac{{BD}}{{CD}} = \frac{{BA}}{{CA}}\end{align}. Prove that $$AD$$ is the bisector of $$\angle BAC$$. ## Text Solution

#### Reasoning:

(i) As we know that in an isosceles triangle, the angles opposite to equal sides are equal.
(ii) Converse of BPT

#### Steps:

Extended $$BA$$ to $$E$$ such that $$AE = AC$$ and join $$CE.$$

In $$\Delta AEC$$

$$AE{\text{ }} = {\text{ }}AC\;\; \Rightarrow \angle ACE = \angle AEC .... {\text{ }}\left( i \right)$$

It is given that

\begin{align} &\frac{{BD}}{{CD}} = \frac{{BA}}{{CA}}\\ &\frac{{BD}}{{CD}} = \frac{{BA}}{{AE}}\left(\because {AC = AE} \right)\_\_\_\_\_\_{\text{ }}\left( {ii} \right) \end{align}

In $$\Delta ABD$$ and $$\Delta EBC$$

\begin{align} &AD||EC\,\,\left( {{\text{Converse of BPT}}} \right)\\ & \Rightarrow \angle BAD = \angle BEC\,\,\left( {{\;\;\;\;\;\;\text{Corresponding angles}}} \right){\text{ }}\_\_\_\_\_{\text{ }}\left( {iii} \right)\\ & and\,\,\angle DAC = \angle ACE\,\,\left( {{\;\;\;\text{Alternative Angles}}} \right){\text{ }}\_\_\_\_\_\_\_\_{\text{ }}\left( {iv} \right) \end{align}

From (i), (iii) and (iv)

\begin{align} &\angle BAD = \angle DAC\\ & \Rightarrow AD\,\text{is the bisector of}\;\angle BAC \end{align}

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