# Circum-centre of triangle formed by external bisectors of base angles of a given triangle is collinear with the other vertices of the two triangles.

So I was solving this question

In $\triangle ABC$, the external bisectors of $\angle A$ and $\angle B$ meet at the point $D$. Prove that the circum-centre of $\triangle ABD$ and the points $C,D$ are collinear.

I made the diagram on GeoGebra and two things popped out - The line $CD$ passes through the incentre $I$ of $\triangle ABC$ and even the circumcircle of $\triangle ABD$ passes through $I$. This essentially solves the question, as $m\angle IAD = 90^{\circ}$, which makes $ID$ the diameter of the circum-circle.

My question here is how do I prove that $AD$ is an angle bisector and that the circum-circle of $\triangle ABD$ passes through $I$.

First, you prove that external bisectors $AD$, $BD$, and inner bisector $CD$ intersect in on point in the same manner as with 3 inner bisectors. => Points $C$, $I$ and $D$ lie on the same line.
Quadrilateral $IADB$ has two opposite angles $\angle IAD=\angle IBD=90^\circ$, so $IABD$ is cyclic with the center of circle $O$ in the center of $ID$. => Points $O$, $I$ and $D$ lie on the same line.