Question

The given figure shows a pentagon $ABCDE$ inscribed in a circle with centre $O$. Given $AB=BC=CD$ and $\angle ABC={132}^{\circ }$. Find $\angle AEB$

Answer 1

Join $BE$

Now, $ABCE$ is a cyclic quadrilateral. [As all vertex lies on the circumference.]

$\therefore ,\angle ABC+\angle AEC={180}^{\circ }$

$\Rightarrow \angle AEC={180}^{\circ }-{132}^{\circ }={48}^{\circ }$

Since, $AB=BC$, therefore,

$\angle AEB=\angle BEC$ [Equal chords subtend equal angles]

$\therefore \angle AEB=\frac{1}{2}\angle AEC$

$=\frac{1}{2}\times {48}^{\circ }$

$\therefore \angle AEB={24}^{\circ }$