Ex. 6.6 Q9 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.6'


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



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


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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