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Let $A B C D$ be a quadrilateral such that there exists a point $E$ inside the quadrilateral satisfying $A E=B E=C E=D E$. Suppose $\angle D A B, \angle A B C, \angle B C D$ is an arithmetic progression. Then the median of the set $\{\angle D A B, \angle A B C, \angle B C D\}$ is
$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
Solution
(d)
Since, $\angle D A B, \angle A B C$ and $\angle B C D$ are in $AP \therefore$ Let $\angle D A B=\theta-\alpha, \angle A B C=\theta$ and $\angle B C D=\theta+\alpha$
$\therefore$ Median of $\angle D A B, \angle A B C$ and $\angle B C D=\theta$
From point $E$ all the vertices are at equal distance.
$\therefore A B C D$ is cyclic.
and $\angle A D C=2 \pi-(\theta-\alpha+\theta+\theta+\alpha)$
$\quad=2 \pi-3 \theta$
and $\angle A D C+\angle A B C=\pi$
$\Rightarrow 2 \pi-3 \theta+\theta=\pi$
$\therefore \quad \theta=\frac{\pi}{2}$
