# Proving line is tangent to circle.

Let $$W_1$$ be a circumcircle of triangle $$ABC$$. $$D$$ be any point on segment $$AC$$. And $$W_2$$ be a circle which is tangent to $$BD$$, $$AD$$ and circle $$W_1$$. $$M$$ be a tangent point on $$AD$$. Then prove that the line parallel to $$BD$$ that passes through the point $$M$$ is tangent to incircle of $$ABC$$. My try: If we take the point which is the intersect of 2 circles as point $$E$$. With homotethy we achieve that $$F$$ is the midpoint of the arc $$AC$$. (Whereas $$F$$ is intersection of $$W_1$$ and $$EM$$). So $$BF$$ is angle bisector of angle $$ABC$$. And if we take the line that is parallel to $$BD$$ as $$l$$. Intersection of $$l$$ and $$AB$$ is $$K$$. Since angle $$DNM$$ $$DMN$$ and $$KMN$$ are equal. $$MN$$ is angle bisector of $$KMD$$. (WHEREAS $$N$$ is tangent point on $$BD$$).Now if we can prove angle bisector of $$BCA$$ or $$BAC$$ passes through the point where $$BF$$ and $$MN$$ intersected we will achieve that quadrilateral $$BKMC$$ is tangential one.

It's another way to phrase Sawayama's Lemma (statement and hints here). It states that if $$N$$ is the tangency point of the yellow circle to $$BD$$ then $$I$$, $$M$$ and $$N$$ are collinear. But $$\triangle MDN$$ is isosceles so $$\angle DMI = \frac{1}{2} \angle CDB$$. You already know that $$MD$$ is one tangent from $$M$$ to the incircle, so the other one must be symmetrical to it about $$MI$$, hence form the angle $$2 \angle DMI = \angle CDB$$ with $$AC$$, implying that it is indeed parallel to $$BD$$.