, 2021

Let $\Gamma$ be the circumcircle of $ABC$, $\Gamma_I$ be the incircle and $\Gamma_{I_a}$ be the A-excircle. Let $\Gamma_1$ and $\Gamma_2$ be the two circles that are tangent to $AB, AC$ and $\Gamma$, where $\Gamma_1$ is inside $\Gamma$ and $\Gamma_2$ is outside $\Gamma$. Let $\Gamma_1$ be tangent to $\Gamma$ at $P$ and $\Gamma_2$ be tangent to $\Gamma$ at $Q$. The key observation is that $P, Q, T, U$ are collinear. This can be split into two parts.

*Lemma 1. $P, Q, U$ are collinear.*
Proof. Note that $P$ is the external homothetic center of $\Gamma$ and $\Gamma_1$, and $Q$ is the internal homothetic center of $\Gamma$ and $\Gamma_2$. Therefore, by Monge theorem, $PQ$ passes the internal homothetic center of $\Gamma_1$ and $\Gamma_2$.

Note that $XI/XI_a = AI/AI_a$ is the ratio of the radii of $\Gamma_I$ and $\Gamma_{I_a}$. Therefore $X$ is the internal homothetic center of $\Gamma_I$ and $\Gamma_{I_a}$. Also note that if we consider the homothety at $A$ with ratio $AU/AX$, then $I$ is sent to the center of $\Gamma_1$, and $I_a$ is sent to the center of $\Gamma_2$ (this is because that if $\Gamma_1$ is tangent to $AB$ and $AC$ at $M$ and $N$, respectively, then it is well-known that $M, I, N$ are collinear, which shows that $\triangle MI$ (center of $\Gamma_1$) ~ $\triangle ZXU$, and a similar argument works for $\Gamma_2$). This shows that $\Gamma_I$ is sent to $\Gamma_1$ and $\Gamma_{I_a}$ is sent to $\Gamma_2$. As $X$ is sent to $U$, we know that $U$ is the internal homothetic center of $\Gamma_1$ and $\Gamma_2$, and thus is passed by $PQ$. $\square$

*Lemma 2. $P, Q, T$ are collinear.*
There are several proofs for this lemma. We present them all here.

Proof. Take the transformation that inverts with respect to $A$ with radius $\sqrt{AB \cdot AC}$ and then reflects with respect to $AI_a$. Then $B \to C$ and $C \to B$. We also have $\Gamma_1 \to \Gamma_{I_a}$ and $\Gamma_2 \to \Gamma_I$. Now suppose that $X \to X'$, $P \to P'$, and $Q \to Q'$, then we know that $\Gamma_I$ is tangent to $BC$ at $Q$ and $\Gamma_{I_a}$ is tangent to $BC$ at $P$. Moreover, we know that $X'$ is on $\Gamma$ because $X$ is on $BC$, and we have $\angle CAX' = \angle XAB = \angle ACB$, showing that $AX'$ is parallel to $BC$. Now we just need to show that $AX'P'Q'$ are concyclic. This is true as $BP' = CQ'$, which shows that $AX'P'Q'$ is an isosceles trapezoid. $\square$

Alternative proof of Lemma 2. We still consider the same transformation as in the proof above. Let the tangents at $P$, $Q$ to $\Gamma$ intersect at $R$. Then it suffices to show that $R$ lies on the polar line of $T$ with respect to $\Gamma$, which is equivalent to showing that if $AR$ intersects $\Gamma$ again at $M$, then $ABMC$ is harmonic. Equivalently, it suffices to show that after the transformation, $\infty CM'B$ is harmonic on the line $BC$, which is equivalent to saying that $M'$ is the midpoint of $BC$. Now note that $M'$ is the midpoint of $P'Q'$ as $APMQ$ is harmonic. Since $BP' = CQ'$, we are done. $\square$

Yet another proof of Lemma 2. We still consider the points $P', Q'$. Let $AP', AQ'$ intersect $\Gamma$ again at $P'', Q''$, respectively. Then it is clear that $PP'', QQ''$ are parallel to $BC$, and to show that $PQ$ passes $T$, it suffices to show that $P''Q'' \cap BC$ is symmetric to $T$ with respect to the midpoint $M$ of $BC$. This is true by the butterfly theorem as $AP'' \cap BC = P'$, $AQ'' \cap BC = Q'$ are symmetric with respect to $M$ and $AA \cap BC = T$. $\square$

Now by the two lemmas, it is clear that $V$ is either $P$ or $Q$. We will next show that it has to be $P$. Note that since $AB < AC$, we know that $TB < TC$ (because $TB/TC = (AB/AC)^2$). We also have $\angle C < \angle B$, showing that $\angle AXB = 90^\circ + \frac{1}{2}(\angle B - \angle C) > 90^\circ$. This shows that $D$ is between $X$ and $C$, and so $\angle DAC < \frac{1}{2}\angle A$. Since it is well-known that $\angle BAP = \angle DAC$, we know that $\angle BAP < \frac{1}{2}\angle A = \angle BAU$. Similarly, we have $\angle BAQ > \angle BAU$. Therefore $T, P, U, Q$ appear in order on the line they are on, and so $V = P$, as desired. As a consequence, $\angle BAV = \angle BAP = \angle DAC$.