The Angle problem

Hi... I know it may be late... but still here is why the quadrilateral is cyclic. I'll take trigonometric approach.

First off let's split the quadrilateral into triangles $\triangle ABD$ and $\triangle ACD$ .

Applying sine rule on both triangles, we observe that $\frac { |BD| }{ \sin { \angle BAD } } =\frac { |AD| }{ \sin { \angle ABD } }$ and $\frac { |DC| }{ \sin { \angle CAD } } =\frac { |AD| }{ \sin { \angle ACD } }$ . However since $\angle BAD=\angle CAD$ and $|BD|=|CD|$ , we can conclude that $\frac{|AD|}{\sin{\angle ABD}}=\frac{|AD|}{\sin{\angle ACD}}$ .

Now we see that the two triangles $\triangle ABD$ and $\triangle ACD$ are never congruent as two of their side lengths are the same and $|AB|\neq |AC|$ . Also $\angle BAD=\angle CAD$ which means that there are no more equal angles in these two triangles, as if one pair of the angles are equal, the other angle pair will be equal also and the triangles will be congruent by AAS.

What this means is that relating back to $\frac{|AD|}{\sin{\angle ABD}}=\frac{|AD|}{\sin{\angle ACD}}$ , we also know that $\sin{\theta}=\sin({180^{o}-\theta})$ . So $\angle ABD+\angle ACD=\angle ABD+ (180^{o}-\angle ABD)=180^{o}$ , hence the quadrilateral is cyclic as opposite angles add up to $180^{o}$ .