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.
Best Answer
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 $BD$.