$AC$ and $BD$ are chords of a circle which bisect each other. Prove that $AC$ and $BD$ are diameters.

(N/A) Let the circle have its center at some point $P$. However,the problem states that the chords $AC$ and $BD$ bisect each other at point $O$.
In $\Delta AOB$ and $\Delta COD$:
$AO = OC$ (Given,as $O$ is the midpoint of $AC$)
$BO = OD$ (Given,as $O$ is the midpoint of $BD$)
$\angle AOB = \angle COD$ (Vertically opposite angles)
By $SAS$ congruence criterion,$\Delta AOB \cong \Delta COD$.
Therefore,$AB = CD$ (Corresponding parts of congruent triangles).
Since $AB = CD$,the arcs corresponding to these chords are equal: $\text{arc } AB = \text{arc } CD$.
Similarly,$\Delta AOD \cong \Delta COB$ by $SAS$ congruence,which implies $AD = CB$,so $\text{arc } AD = \text{arc } BC$.
Now,consider the quadrilateral $ABCD$. Since the diagonals $AC$ and $BD$ bisect each other,$ABCD$ is a parallelogram.
In a parallelogram inscribed in a circle,opposite angles are equal ($\angle A = \angle C$ and $\angle B = \angle D$). Since the sum of opposite angles in a cyclic quadrilateral is $180^{\circ}$,we have $\angle A + \angle C = 180^{\circ}$,which implies $2\angle A = 180^{\circ}$,so $\angle A = 90^{\circ}$.
Since $\angle A = 90^{\circ}$,the chord $BD$ subtends a right angle at the circumference,which means $BD$ must be a diameter. Similarly,$AC$ is a diameter.