Let $ABC$ be a triangle with $AB \ne AC$. The incirle of triangle $ABC$ is tangent to $BC, CA, AB$ at $D, E, F$, respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$. The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX \perp T F$
Let $M$ be the midpoint of arc $BC$ different than arc $BAC$, $N$ - the intersection of $AM$ and $BC$, $I$ - the incenter of triangle $ABC$. By inscribed angles theorem
$$\triangle ABN\sim\triangle AMC\implies \frac{AB}{BN}=\frac{AM}{CM}.$$Trilium theorem yields $MI=CM$. Since $AN\perp EF\perp DX$ we have $$\frac{BX}{BD}=\frac{AB}{BN}=\frac{AM}{MI}.$$Tangent segments theorem gives $BD=BF$. Hence $$\frac{BX}{AM}=\frac{BF}{MI}.$$We have $\angle AEI=90^\circ=\angle AFI$ thus segment $AI$ is the diameter of circle $AEF$ and $\angle ATI=90^\circ$. Hence $$\measuredangle TFB=\measuredangle TFA=\measuredangle TIA=\measuredangle TIM.$$Additionally $$\measuredangle TBF=\measuredangle TBA=\measuredangle TMA=\measuredangle TMI$$thus $$\triangle TFB\sim\triangle TIM\implies \frac{BT}{MT}=\frac{BF}{MI}=\frac{BX}{AM}.$$Because $\measuredangle TBX=\measuredangle TMA$ we have from $(s,a,s)$
$$\triangle XTB\sim\triangle ATM.$$Two last similarities imply$$\triangle TXF\sim\triangle TAI\implies \angle XTF=\angle ATI=90^\circ.$$QED
#1673