Post Number:#2 by ahmedittihad » Sun Jan 01, 2017 4:00 am
We will show the result by the** radical axis theorem**. So, we will show that $BFDX, BMDE, FEMX$ is concyclic.
$ \angle ADF = \angle ACB$ by **spiral similarity**.
$\angle FDC = 180 - \angle ADE - \angle CDX = 180 - 2\angle ADE - \angle ADF = 180 - \angle BFC - \angle ACB = 90 $.
So, $BCDF$ is a cyclic quadrilateral and $M$ is the center of this circle. So, $\angle AMD = 2\angle ACD = \angle MAE$. Along with the fact that $ AC \parallel DE$ we get that $AMDE$ is an isosceles trapizium. Meaning also that it is cyclic.
Now, $ \angle BAD + \angle BMD = 2\angle BAC + 2\angle ACD + 2\angle BCF = 2( 2\angle BAC + \angle BCF) = 2(\angle BFC + \angle BCF) = 180 $. So,$ABMD$ is cyclic. Also, $ \angle ABD = \angle ABF + \angle FBD = \angle ABF + \angle ACD = 2\angle BAC$. So, $ \angle ABD + \angle AED = 2\angle BAC +180 - 2\angle DAE = 180$.
So we get, $ABMDE$ is cyclic.
Now, we prove that $BDFX$ is cyclic. $\angle AMD = \angle EAM = \angle EXM = \angle MDX$. So, $\triangle MDX$ is isosceles. So, $X$ lies on the circle $BFDC$.
At last, we need to prove that $FEXM$ is cyclic. See that. $A$ is the center of similitude that sends $BD$ to $FE$. So, $ \angle ABD = \angle AFE = 2\angle BAC$.
We already have that $\angle MXE = 2\angle BAC$. And the result follows.
Frankly, my dear, I don't give a damn.