Prove that the mid-point of the hypotenuse of a right angled triangle is equidistant from its vertices.

In the given figure, what will be the value of $\angle CBA?$

In the given figure, $O$ is the centre of the circle anc $\angle AOD=106°$, what will be the value of $\angle BCD$?

In the adjoining figure, $OPQR$ is a square, A circle drawn with center $O$ cuts the square in $X$ and $Y$. Then, what will be the relation between $QX \mathrm{and} QY$.