Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.

$ABCD$ is a rectangle. Prove that the center of the circle through $A, B, C, D$ is the point of intersection of its diagonals.

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the center of the circle. Justify the construction.

In a cyclic quadrilateral $ABCD$, if $\left(\angle B-\angle D\right)=60°$, show that the smaller of the two is $60°$.

The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

On a common hypotenuse $AB,$ two right triangles $ACB$ and $ADB$ are situated on opposite sides. Prove that $\angle BAC=\angle BDC.$