Prove that the centre of the circle circumscribing the cyclic rectangle $ABCD$ is the point of intersection of its diagonals.

$ABCD$ is a cyclic quadrilateral in which $BA$ and $CD$ when produced meet in $E$ and $EA=ED$ Prove that:

$AD\Vert BC$

$ABCD$ is a cyclic quadrilateral in which $BA$ and $CD$ when produced meet in $E$ and $EA=ED$ Prove that:

$EB=EC$

In the given figure, two circles intersect at $A$ and $B$. The centre of the smaller circle is $O$ and it lies on the circumference of the larger circle. If $\angle APB=70°$, find $\angle ACB$.

In the given figure, two congruent circles with centres $O$ and $O$' intersect at $A$ and $B$. If $\angle A{O}^{\text{'}}B=50°$ then find $\angle APB$.

In the given figure, $ABCD$ is a cyclic quadrilateral in which $\angle BAD=75°, \angle ABD=58°$ and $\angle ADC=77°$, $AC$ and $BD$ intersect at $P$. Then, find $\angle DPC$.