The bisector of the exterior angle at vertex C of triangle ABC intersects the circumscribed circle at point D. Prove that AD=BD

Question

The bisector of the exterior angle at vertex $C$ of triangle $ABC$ intersects the circumscribed circle at point $D$. Prove that $AD=BD$.

So what I'm wondering is how to prove this? I've already drawn a diagram but I don't know how to continue from there. Please help!

Answer 1

Because $$\measuredangle DCB=180^{\circ}-\measuredangle ACD=\measuredangle ABD,$$ which says $$AD=BD.$$

Answer 2

Note that since $CD$ is the external bisector of $\angle ACB$ $ \implies \angle BCA = 180-2\angle DCB \implies \angle ACD= 180-BCD \implies \angle ABD = \angle BCD $ ( using sum of opposite angles is equal to 180 ) .

Again by cyclic quads, we get that $\angle BCD= \angle DAB$.

Hence we have $\angle DAB=\angle BCD=\angle ABD$ and the result follows.