Prove that equal chords of congruent circles subtend equal angles at their centre.

Given two equal chords $AB$ and $CD$ of a circle, with centre $O$, intersecting each other at point $P$. Prove that $\mathrm{AP}=\mathrm{CP}$.

Given two equal chords $AB$ and $CD$ of a circle, with centre $O$, intersecting each other at point $P$. Prove that $\mathrm{BP}=\mathrm{DP}$.

In the given figure, $O$ is the centre of the circle with radius $20 \mathrm{cm}$ and $\mathrm{OD}$ is perpendicular to $AB$. $\mathrm{AB}=32 \mathrm{cm}$. If the length of $CD$ is $k \mathrm{cm}$, find $k$.

In the given figure, $AB$ and $CD$ are two equal chords of a circle, with centre $O$. $P$ is the mid-point of chord $AB, Q$ is the mid-point of chord $CD$ and $\angle \mathrm{POQ}=150°$. If $\angle \mathrm{APQ}=k°$, find $k$.