Problem 1250

Define $A''$, $B''$, $C''$ to be the intersection points of the lines $AA'$, $BB'$, $CC'$ with the sides $BC$, $CA$, $AB$, respectively, $B_a$, $C_a$ the intersection points of the tangent line to $\omega$ at $A'$ with the lines $AB$, $AC$, respectively, $C_b$, $A_b$ the intersection points of the tangent line to $\omega$ at $B'$ with the lines $BC$, $BA$, respectively, $A_c$, $B_c$ the intersection points of the tangent line to $\omega$ at $C'$ with the lines $CA$, $CB$, respectively, $A_1$, $B_1$, $C_1$ the intersection points of the three tangent lines to $\omega$ at $A'$, $B'$, $C'$, as shown in the figure, $r$ the radius of $\omega$ and $r_0$ the inradius of triangle $ABC$.
The distance from $A_1$ to each line $AB$, $AC$ is equal to $|r - r_0|$. Therefore, $A_1$ is on the bisector of angle $\angle BAC$. But $AA_bA_1A_c$ is a parallelogram. We deduce that $AA_bA_1A_c$ is a rhombus and therefore $A_1A_b = A_1A_c$. On the other hand, we have $A_1B' = A_1C'$ as tangents to $\omega$ from $A_1$. We deduce that $A_bB' = A_cC'$. We deduce in a similar way that $B_cC' = B_aA'$ and $C_aA' = C_bB'$.

Because lines $B_aC_a$ and $BC$ are parallel, we have from Thales
$$\frac{BA''}{A''C} = \frac{B_aA'}{A'C_a}.$$
We have similarly
$$\frac{CB''}{B''A} = \frac{C_bB'}{B'A_b},$$
and
$$\frac{AC''}{C''B} = \frac{A_cC'}{C'B_c}.$$
Therefore
$$\frac{BA''}{A''C} \cdot \frac{CB''}{B''A} \cdot \frac{AC''}{C''B} = \frac{B_aA'}{A'C_a} \cdot \frac{C_bB'}{B'A_b} \cdot \frac{A_cC'}{C'B_c} = 1,$$
and by Ceva's theorem, lines $AA'$, $BB'$, $CC'$ are concurrent.

Because $IA'$ and $IC'$ are perpendicular to $BC$ and $AB$, respectively, at the intouch points of the incircle of $ABC$, and $IA' = IC'$, we have $\angle CBA' = \angle C'BA$ and $\angle A'BA = \angle CBC'$ as oriented angles. We have in a similar way $\angle BAC' = \angle B'AC$, $\angle C'AC = \angle BAB'$, $\angle ACB' = \angle A'CB$ and $\angle B'CB = \angle ACA'$ as oriented angles.
Because $AA'$, $BA'$, $CA'$ intersect in a point $A'$, we have from the trigonometric form of Ceva's theorem
$$\frac{\sin \angle BAA'}{\sin \angle A'AC} = \frac{\sin \angle A'BA}{\sin \angle CBA'} \cdot \frac{\sin \angle A'CB}{\sin \angle ACA'}.$$
Similarly, we have
$$\frac{\sin \angle CBB'}{\sin \angle B'BA} = \frac{\sin \angle B'CB}{\sin \angle ACB'} \cdot \frac{\sin \angle B'AC}{\sin \angle BAB'},$$
and
$$\frac{\sin \angle ACC'}{\sin \angle C'CB} = \frac{\sin \angle C'AC}{\sin \angle BAC'} \cdot \frac{\sin \angle C'BA}{\sin \angle CBC'}.$$

After multiplication and cancellation, we obtain
$$\frac{\sin \angle BAA'}{\sin \angle A'AC} \cdot \frac{\sin \angle CBB'}{\sin \angle B'BA} \cdot \frac{\sin \angle ACC'}{\sin \angle C'CB} = 1.$$
We deduce from the trigonometric form of Ceva's theorem that $AA'$, $BB'$, $CC'$ are concurrent.