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(D) Since $AE=BE=CE=DE$,the point $E$ is the circumcenter of the quadrilateral $ABCD$. Thus,$ABCD$ is a cyclic quadrilateral.Let the angles be $\angle

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(D) Since $AE=BE=CE=DE$,the point $E$ is the circumcenter of the quadrilateral $ABCD$. Thus,$ABCD$ is a cyclic quadrilateral.Let the angles be $\angle DAB = 	heta - \alpha$,$\angle ABC = 	heta$,and $\angle BCD = 	heta + \alpha$. The median of these angles is $	heta$.In a cyclic quadrilateral,the sum of opposite angles is $\pi$. Thus,$\angle ADC + \angle ABC = \pi$,which means $\angle ADC = \pi - 	heta$.The sum of the interior angles of a quadrilateral is $2\pi$. Therefore,$\angle DAB + \angle ABC + \angle BCD + \angle ADC = 2\pi$.Substituting the values: $(	heta - \alpha) + 	heta + (	heta + \alpha) + (\pi - 	heta) = 2\pi$.$2	heta + \pi = 2\pi$.$2	heta = \pi \implies 	heta = \frac{\pi}{2}$.

Let $ABCD$ be a quadrilateral such that there exists a point $E$ inside the quadrilateral satisfying $AE=BE=CE=DE$. Suppose $\angle DAB, \angle ABC, \angle BCD$ are in an arithmetic progression. Then the median of the set $\{\angle DAB, \angle ABC, \angle BCD\}$ is

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