The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Prove it.

In the figure $\mathrm{O}\mathrm{D}$ is perpendicular to the chord $\mathrm{A}\mathrm{B}$ of a circle with centre $\mathrm{O}$. If $\mathrm{B}\mathrm{C}$ is a diameter show that $\mathrm{A}\mathrm{C}\parallel \mathrm{O}\mathrm{D}$ and $\mathrm{A}\mathrm{C}=2\mathrm{O}\mathrm{D}$.

In the figure $\mathrm{A}\mathrm{B}$ is a diameter of the circle $\mathrm{C}\mathrm{D}$ is a chord equal to radius of the circle. $\mathrm{A}\mathrm{C}\mathrm{}$and $\mathrm{B}\mathrm{D}$ when produced meet at $\mathrm{E}$. Prove that $\mathrm{\angle }\mathrm{A}\mathrm{E}\mathrm{B}=60°$.

In the figure, $\mathrm{ABCD}$ is a cyclic quadrilateral such that $\mathrm{AC} \mathrm{and} \mathrm{BD}$ are its diagonals. If $\angle \mathrm{DBC}=55° \mathrm{and} \angle \mathrm{BAC}=45°$ and $\angle \mathrm{BCD}=k°$, then find $k$.