Post Number:#2 by tafhim01 » Thu Jan 19, 2017 6:46 pm
[AST: Alternate Segment Theorem]
We assume that, WLOG radius of $\omega$ is less than radius of $\omega'$(equality case is trivial).
Let $P$ be the midpoint of $\widehat{EF}$ not containing $B$ of $\odot BEF$, thus $BP$ is the angle bisector of $\angle EBF$. Let the tangent at $P$ to $BEF$ intersect $XY$ at $M$.
By AST and similarity,
\[\angle ABD = \angle ABC = \angle B.\]
Now,
\[\angle MPE = \angle PFE = \angle PEF \Rightarrow PM \parallel EF.\]
Now,
\[EF \perp XY \Rightarrow PM \perp XY.\]
Extend $\overrightarrow{CA}$ to $C'$.
\[\angle DAX = \angle C'AX \text{ [External angle bisector]}\]
Again, by AST,
\[\angle ABX = \angle C'AX.\]
So,
\[\begin{align*}
&\angle ABX = \angle DAX = \angle DBX \\
&\\
\Rightarrow &X\text{ is the midpoint of }\widehat{AD}\text{ not containing B.}\\
&\text{Similarly, }Y\text{ is the midpoint of arc }AC.
\end{align*}\]
Now,
\[\begin{align*}
&\angle ABE = \angle ADE = \angle CAF = \angle CBF\\
&\\
\Rightarrow &\angle EBF = \angle ABC - \angle ABE + \angle CBF = \angle ABC = \angle B.
\end{align*}\]
So,
\[\angle EFP = \frac{\angle B}{2}\]
But,
\[\angle AFC = \angle ABC = \angle B.\]
So, $PF$ bisects $\angle AFC \Rightarrow Y, P, F$ are collinear. Similarly, $P, E, X$ are collinear.
By simple angle chasing,
\[
\begin{align*}
&\triangle ADE \sim \triangle CAF\\
&\\
\Rightarrow &\angle ACF = \angle ADE.
\end{align*}
\]
Now,
\[
\begin{align*}
&\angle PYX = \angle ACF = \angle ADE = \angle PXY.\\
&\\
\Rightarrow &\triangle PXY \text{ is isosceles.}\\
\Rightarrow &PM \text{ is the perpendicular bisector of } XY\text{ , and also tangent to } BEF\text{ at }P.
\begin{align*}
&\\
&\\
&\square
\end{align*}
\end{align*}
\]