A self-made geo

[rubab]

ABC be a triangle with circumcircle $\Omega$. Let the angle bisector of $\angle BAC$ intersect BC at x. Let the incircle touches BC, AC, and AB at D,E,F respectively.Prove that $\Omega$, $\bigodot ADX$, $\bigodot AEF$ are coaxal.

[Raiyan Jamil]

Solution:

Let $A'$ be the midpoint of the arc $BC$ not containing $A$. And let $A'D$ intersect circumcircle of $ABC$ at $P$. From $Canada- 2007/5$, we get that $APEF$ is a cyclic quadrilateral. We now need to prove that $APDX$ is cyclic which can be done easily by angle chasing.

[rubab]

Solution:

apply an inversion centred A that takes the incircle to the A-excircle (let it be $\alpha$).we call the inversion $\sigma$.We define $P'$ as the inversion of any point, $P$.
$\sigma$ takes BC to $\bigodot AB'C'$ which is tangent to $\alpha$.Obviously The point of tangency is $D$' .So, $\alpha$ is A-mixtillinear incircle of triangle $AB'C'$.now $X'$ is the midpoint of arc $B'C'$ in $\bigodot AB'C'$ not containing $A$. And $E',F'$ are the point of tangency of $\alpha$ with $AC'$ and $AB'$ respectively.so $\sigma$ takes $\omega$ to $B'C'$,$\bigodot ADX$ to D'X',$\bigodot AEF$ to $E'F'$."By E chen's lemma $"B'C', D'X', E'F'$ are concurrent.

[tanmoy]

Let $I$ be the incenter of $\triangle ABC$ and let $AI \cap (ABC)=E_{1},E_{1}D \cap (ABC)=F_{1}$.

Lemma 1: $A,X,D,F_{1}$ are cyclic.
Proof: $\measuredangle E_{1}BD=\measuredangle BF_{1}D$ $\Rightarrow$ $E_{1}D \cdot E_{1}F_{1}=E_{1}B^{2}=E_{1}X \cdot E_{1}A$.So,the lemma is proved.

Lemma 2: $A,E,F,F_{1}$ are cyclic.
Proof: From lemma 1,$E_{1}D \cdot E_{1}F_{1}=E_{1}B^{2}=E_{1}I^{2}$ $\Rightarrow$ $\measuredangle E_{1}ID=\measuredangle DF_{1}I$...............(i).We also have that
$\measuredangle AF_{1}E_{1}=\measuredangle AF_{1}I + \measuredangle DF_{1}I=\measuredangle ABE_{1}=\measuredangle E_{1}XB=\measuredangle IDX + \measuredangle E_{1}ID$.So,from (i),$\measuredangle AF_{1}I=\measuredangle IDX= \dfrac {\pi} {2}=\measuredangle AFI=\measuredangle AEI$.
Thus the lemma is proved.

From Lemma 1 and Lemma 2,we can conclude that $AF_{1}$ is the common pairwise radical axis of $\Omega,(AEF)$ and $(ADX)$.So,they are coaxal.