Question

In the figure, $AD$ is the internal bisector of $\angle A$ and $CE\parallel DA$. If $CE$ meets $BA$ produced at $E$, prove that $\Delta CAE$ is isosceles.

Answer 1

Step-by-step explanation:

In the given figure $AD$ is internal bisector of angle $A$ and $CE$ is parallel to $DA$. If $CE$ meets $BA$ produced at $E$ prove that triangle $CAE$ is isosceles

$\angle BAC+\angle EAC=180°$ ( Straight line)

In $\Delta ACE$,

$\angle ACE+\angle AEC+\angle EAC=180°$ ( sum of angles of Triangle)

Equating both we get,

$\angle BAC+\angle EAC=\angle ACE+\angle AEC+\angle EAC$

$=>\angle BAC=\angle ACE+\angle AEC$ .......... [Eq 1]

$\angle BAD=\left(\frac{1}{2}\right)\angle BAC$

$\angle BAD=\angle AEC$....... $\left(AD║CE\right)$

$=>\angle AEC=\left(\frac{1}{2}\right)\angle BAC$

Putting this in eq (1) we get,

$=>\angle BAC=\angle ACE+\left(\frac{1}{2}\right)\angle BAC$

$=>\angle ACE=\left(\frac{1}{2}\right)\angle BAC$

$\angle AEC=\angle ACE=\left(\frac{1}{2}\right)\angle BAC$

$=>AC=AE$ {Converse of isosceles triangle having equal sides with equal angles property}

Hence $\Delta CAE$ is an isosceles triangle.