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Question

(d)

Since, $\angle D A B, \angle A B C$ and $\angle B C D$ are in $AP 	herefore$ Let $\angle D A B=	heta-\alpha, \angle A B C=	heta$ and $\angle

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(d)

Since, $\angle D A B, \angle A B C$ and $\angle B C D$ are in $AP 	herefore$ Let $\angle D A B=	heta-\alpha, \angle A B C=	heta$ and $\angle B C D=	heta+\alpha$

$	herefore$ Median of $\angle D A B, \angle A B C$ and $\angle B C D=	heta$

From point $E$ all the vertices are at equal distance.

$	herefore A B C D$ is cyclic.

and $\angle A D C=2 \pi-(	heta-\alpha+	heta+	heta+\alpha)$

$\quad=2 \pi-3 	heta$

and $\angle A D C+\angle A B C=\pi$

$\Rightarrow 2 \pi-3 	heta+	heta=\pi$

$	herefore \quad 	heta=\frac{\pi}{2}$

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

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