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!
Best Answer
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.