Question

In the given figure, $AB$ and $CD$ are two parallel chords of a circle. if $BDE$ and $ACE$ are straight lines, intersecting at $E$, prove that $\Delta AEB$ is isosceles.

Answer 1

$AB$ and $CD$ are parallel chords of circle. Line $BD$ and $AC$ intersect at point $E$.

As we know that in a cyclic quadrilateral, the exterior angle is equal to opposite interior angle then

Exterior angle $\angle CDE=\angle A,$ ......(1)

Exterior angle $\angle ECD=\angle B$ ......(2)

Now, $AB\parallel CD$ then

$\angle A=\angle ECD$ when $AE$ is tranversal,

Hence,

$\angle A=\angle B.$ [By equation (2)]

Also, we know in a triangle, opposite sides of equal angles are equal then

$AE=BE$ (in $△AEB$ )

Hence, $△AEB$ is isosceles.