Ex. 6.6 Q9 Triangles Solution - NCERT Maths Class 10

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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 \(\angle BAC\).

 Video Solution
Ex 6.6 | Question 9

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 \(\Delta AEC\)

\[ \begin{align}AE &= AC \\ \Rightarrow \angle ACE &= \angle AEC \;\;\dots\left(\text{i} \right) \end{align} \]

It is given that

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

In \(\Delta ABD\) and \(\Delta EBC\)

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

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

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

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