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 $A^{\prime }$ be the midpoint of $A_1{A}_2$. $B^{\prime }$ and $C^{\prime }$ are defined similarly. $A$, $B$ and $C$ are collinear if and only if $A^{\prime }$, $B^{\prime }$ and $C^{\prime }$ are collinear.

Proof. The key observation is that $\overrightarrow{A^{\prime }{B}^{\prime }}=\frac{1}{2}(\overrightarrow{A_1{B}_1}+\overrightarrow{A_2{B}_2})$ and $\overrightarrow{B^{\prime }{C}^{\prime }}=\frac{1}{2}(\overrightarrow{B_1{C}_1}+\overrightarrow{B_2{C}_2})$. Using these observations and the Thales' theorem, it is easy to check that the assertion is equivalent to the equality $\frac{A_1{B}_1}{A_2{B}_2}=\frac{B_1{C}_1}{B_2{C}_2}$. $\square$

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 $T^{\prime }$ is the midpoint of $H_b{H}_c$, then $T^{\prime }$ lies on the perpendicular bisector of side $BC$.

Proof. Let $N$ be the midpoint of side $BC$. Note that $T{H}_c\perp C{H}_c$ and $TN\perp BC$, hence the quadrilateral $T{H}_{c}NC$ is cyclic and $$\angle H_{c}NB=\angle H_{c}TC=90^{\circ }-\angle TC{H}_c=\angle H_{c}CB+\angle NTC=90^{\circ }-\frac{\angle B}{2}.$$ It means that $H_{c}N$ is parallel to the external bisector of $\angle BAC$. On the other hand, $T{H}_b$ is also parallel to the external bisector of $\angle ABC$ and thus $T{H}_b\parallel H_{c}N$. Similarly, $T{H}_c\parallel H_{b}N$. So $T{H}_{c}N{H}_b$ is a parallelogram and hence $T^{\prime }$, the midpoint of $H_b{H}_c$, 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 $T^{\prime }$ 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 $M^{\prime }$. If $X^{\prime }$ is the image of $X$ under this transformation, then points $A$, $I$, $X^{\prime }$ and $M^{\prime }$ are on a common circle. So $\angle I{X}^{\prime }{M}^{\prime }=\angle IA{M}^{\prime }=90^{\circ }$. On the other hand, since the line $BC$ is the image of the circumcircle of triangle $ABC$ under this transformation, $X^{\prime }$ lies on the line $BC$. This implies that $X^{\prime }=D$ and thus $\angle BAD=\angle CAX$.