Post Number:**#2** by **ahmedittihad** » Sat Apr 01, 2017 2:26 pm

Assume, $AB<AC$.

Reflect $A$ over $X$ and $Y$ to get $K$ and $L$ respectively. We know that the length of the tangent from $A$ to the $A$-excircle is half the perimeter. So, we get, $K$ and $L$ are the touchpoints of the $A$-excircle with $AB$ and $AC$.

Let the $A$-excircle meet $BC$ at $Z$.

Now, the wow factor.

We show that $MZ=AM$.

As $X$ and $Y$ are the midpoints of $AK$ and $AL$, $XY$ is the radical axis of the $A$-excircle and the zero radius circle at $A$. And $M$ lies on the radical axis. Yielding $MZ=AM$.

Now, $AK=2=AB+BK=AB+BZ=AB+BM+MZ=AB+BM+AM$.

The other cases are similar too.

Frankly, my dear, I don't give a damn.