Preselection tests for the full-time training

Let $E$ be the intersection point of $BK$ with $AT$. The problem is equivalent to prove that points $A^{\prime }$, $E$, $B^{\prime }$ are collinear.

First solution. Let $F$ be the intersection point of $BC$ with $KT$. Since $K$ is the midpoint of $\widetilde{CA}$, by expressing the angles in terms of arc lengths we obtain $$\measuredangle KEA=\frac{1}{2}(\text{overparen}KA+\text{overparen}BT)=\frac{1}{2}(\text{overparen}CK+\text{overparen}BT)=\measuredangle CFK=90^{\circ }$$ But $B^{\prime }$ is an intouch point. This implies that $$\measuredangle A{B}^{\prime }I=90^{\circ }=\measuredangle AEI,$$ and the points $A$, $I$, $E$, $B^{\prime }$ are concyclic.



We deduce from this that $$\measuredangle I{B}^{\prime }E=\measuredangle IAE=90^{\circ }-\measuredangle EIA=90^{\circ }-\frac{1}{2}\measuredangle BAC-\frac{1}{2}\measuredangle CBA=\frac{1}{2}\measuredangle ACB.$$ On the other hand, triangle $I{A}^{\prime }{B}^{\prime }$ is isosceles ($I{A}^{\prime }=I{B}^{\prime }$). Therefore, $$\measuredangle I{B}^{\prime }{A}^{\prime }=\frac{1}{2}\left(180^{\circ }-\measuredangle A^{\prime }I{B}^{\prime }\right)=\frac{1}{2}\measuredangle ACB=\measuredangle I{B}^{\prime }E,$$ since $\measuredangle C{A}^{\prime }I=90^{\circ }$. This proves that the points $A^{\prime }$, $E$, $B^{\prime }$ are collinear.

Second solution. We prove as in the first solution that $\measuredangle KEA=90^{\circ }$. Let $D$ be the intersection point of $AT$ with $BC$. The pedal triangle of $I$ with respect to triangle $ADC$ is $A^{\prime }{B}^{\prime }E$. Proving that points $A^{\prime }$, $B^{\prime }$, $E$ are collinear is equivalent to proving that $I$ is on the circumcircle of triangle $ADC$ and therefore, the line $A^{\prime }{B}^{\prime }E$ will be the Simson line of point $I$ with respect to triangle $ADC$.



We have $$\measuredangle CDA=\measuredangle DBA+\measuredangle BAD=\measuredangle CBA+90^{\circ }-\frac{1}{2}\measuredangle CBA=90^{\circ }+\frac{1}{2}\measuredangle CBA.$$ On the other hand $$\measuredangle CIA=180^{\circ }-\frac{1}{2}\measuredangle BAC-\frac{1}{2}\measuredangle ACB=90^{\circ }+\frac{1}{2}\measuredangle CBA=\measuredangle CDA.$$ This proves that $I$ is on the circumcircle of triangle $ADC$ and thus $A^{\prime }$, $E$, $B^{\prime }$ are collinear.