Question

In $\Delta ABC$, the internal bisector of $\angle$ A meets opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Hence $\Delta ACE$ is Right angled triangle.
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given: $△ABC$, AD bisects $\angle BAC$, $CE\parallel DA$ meets AB produced at E
To prove: ACE is an Isosceles triangle
Since, AD bisects $\angle BAC$
$\angle BAD=\angle DAC$
Now, $AD\parallel CE$
$\angle BAD=\angle AEC$ (Corresponding angles)
and, $\angle DAB=\angle ACE$ (Alternate angles)
Thus, $\angle AEC=\angle ACE=\angle BAD=\angle DAC$
In $△ACE$,
$\angle AEC=\angle ACE$
thus, $AC=AE$ (Sides opposite to equal angles are equal)
Hence, $△ACE$ is an Isosceles triangle.