Ex. 6.6 Q9 Triangles Solution - NCERT Maths Class 10

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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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