Problem 2392

Let $E$ be the intersection point of $BK$ with $AT$. The problem is equivalent to prove that points $A'$, $E$, $B'$ 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
$$\text{KEA = 1 2 ( KA + BT ) = 1 2 ( CK + BT ) = CFK = 90}$$
But $B'$ is an intouch point. This implies that
$$\measuredangle AB'I = 90^\circ = \measuredangle AEI,$$
and the points $A$, $I$, $E$, $B'$ are concyclic.

We deduce from this that
$$\measuredangle IB'E = \measuredangle IAE = 90^\circ - \measuredangle EIA = 90^\circ - \frac{1}{2} \measuredangle BAC - \frac{1}{2} \measuredangle CBA = \frac{1}{2} \measuredangle ACB .$$
On the other hand, triangle $IA'B'$ is isosceles ($IA' = IB'$). Therefore,
$$\measuredangle IB'A' = \frac{1}{2}\left(180^\circ - \measuredangle A'I B'\right) = \frac{1}{2} \measuredangle ACB = \measuredangle IB'E,$$
since $\measuredangle CA'I = 90^\circ$. This proves that the points $A'$, $E$, $B'$ 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 $A'B'E$. Proving that points $A'$, $B'$, $E$ are collinear is equivalent to proving that $I$ is on the circumcircle of triangle $ADC$ and therefore, the line $A'B'E$ 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 $A'$, $E$, $B'$ are collinear.