Iranian Mathematical Olympiad

First, we prove two lemmas.

Lemma 1. Let $l_1$ and $l_2$ be two lines and $A$, $B$ and $C$ three points in the plane. Denote by $A_1$ and $A_2$ the perpendicular projections of $A$ on $l_1$ and $l_2$, respectively. Furthermore, let $A'$ be the midpoint of $A_1A_2$. $B'$ and $C'$ are defined similarly. $A$, $B$ and $C$ are collinear if and only if $A'$, $B'$ and $C'$ are collinear.

Proof. The key observation is that $\overrightarrow{A'B'} = \frac{1}{2}(\overrightarrow{A_1B_1} + \overrightarrow{A_2B_2})$ and $\overrightarrow{B'C'} = \frac{1}{2}(\overrightarrow{B_1C_1} + \overrightarrow{B_2C_2})$. Using these observations and the Thales' theorem, it is easy to check that the assertion is equivalent to the equality $\frac{A_1B_1}{A_2B_2} = \frac{B_1C_1}{B_2C_2}$. $\square$

Lemma 2. Let $T$ be the midpoint of arc $BAC$ of the circumcircle of triangle $ABC$. Furthermore, let $H_b$ and $H_c$ be the feet of the perpendicular lines from $T$ to the internal bisectors of $\angle B$ and $\angle C$, respectively. If $T'$ is the midpoint of $H_bH_c$, then $T'$ lies on the perpendicular bisector of side $BC$.

Proof. Let $N$ be the midpoint of side $BC$. Note that $TH_c \perp CH_c$ and $TN \perp BC$, hence the quadrilateral $TH_cNC$ is cyclic and
$$\angle H_c N B = \angle H_c T C = 90^\circ - \angle T C H_c = \angle H_c C B + \angle N T C = 90^\circ - \frac{\angle B}{2}.$$
It means that $H_cN$ is parallel to the external bisector of $\angle BAC$. On the other hand, $TH_b$ is also parallel to the external bisector of $\angle ABC$ and thus $TH_b \parallel H_cN$. Similarly, $TH_c \parallel H_bN$. So $TH_cNH_b$ is a parallelogram and hence $T'$, the midpoint of $H_bH_c$, lies on $TN$. But $TN$ is the perpendicular bisector of side $BC$, which is what we desired to prove.

We will keep using the notations in lemma 2. We will now apply lemma 1, with $IC$, $IB$, $I_a$, $X$ and $M$ playing the roles of $l_1$, $l_2$, $A$, $B$ and $C$, respectively. Since $N$, $M$ and $T'$ are collinear, we obtain that $I_a$, $X$ and $T$ are collinear, too. Now, by an inversion with center $A$ and power $AB \times AC$ and then a reflection with respect to the internal bisector of angle $\angle BAC$, $B$ is sent to $C$, $C$ is sent to $B$, $I_a$ is sent to $I$ and $M$ is sent to the foot of the external bisector of $\angle A$, say $M'$. If $X'$ is the image of $X$ under this transformation, then points $A$, $I$, $X'$ and $M'$ are on a common circle. So $\angle IX'M' = \angle IAM' = 90^\circ$. On the other hand, since the line $BC$ is the image of the circumcircle of triangle $ABC$ under this transformation, $X'$ lies on the line $BC$. This implies that $X' = D$ and thus $\angle BAD = \angle CAX$.