Question

In a $△ABC$, 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 $E$. Then the $△ACE$ is not isosceles.

• A. True
• B. False

Answer 1

The correct option is B False

Given: AD bisects $\angle A$, $CE\parallel AD$
In $△ABC$,
$\frac{BD}{DC}=\frac{AB}{AC}$ (Angle bisector theorem)...(I)
In $△BCE$, $CE\parallel AD$, So,
$\frac{BD}{DC}=\frac{AB}{AE}$ (Side splitter theorem)...(II)
From I and II,
$\frac{AB}{AC}=\frac{AB}{AE}$
hence, $AC=AE$
or $△ACE$ is an isosceles triangle.