Iranian Mathematical Olympiad

We shall prove the following lemma.

Lemma 1. *The points $E$, $F$, $B$, and $C$ are on a circle centered at $T'$.*

Proof. An easy angle chasing shows that $BAEQ$ and $CAFQ$ are indeed cyclic.
Thus,
$$\angle FBE = \angle QBE = \angle QAE = \angle QAF = \angle QCF = \angle FCE.$$
Yielding $EFBC$ is cyclic. Then,
$$\angle BEC = \angle BQE + \angle EBQ = \alpha + \angle FBE = \alpha + \angle QBF = \frac{3\alpha}{2}$$
and
$$\angle BT'C = \angle BAC = 3\alpha = 2\angle BEC$$
And
$$BT' = CT'.$$

Now, we need the following lemma;

Lemma 2. *The quadrilateral $EFA'T$ is cyclic*

Proof. According to the above-mentioned lemma; $\angle FT'E = 2\angle FBE = 2\angle QBF = 2\angle QAE = \angle EAF$. This completes our proof.

Thus the radical axis of $AT'EF$, $AT'CB$, and $EFBC$ are concurrent. Let $G$ be the foot of ex-angle bisector of $\angle BAC$, and assume $BE$ and $CF$ meet at $P'$. Since $BC$ and $GP'$ meet at $X$, it follows that $P'$ is on $AD$, indeed $(GX, BC) = (GD, BC) = -1$. Yielding $X = D$. That is;
$$\angle BP'C = \angle BEC + \angle FCE = \angle BEC + \angle FBE = 2\alpha$$
Yielding to the fact that $P$ is the intersection of the corresponding arcs of $AD$ and $BC$. Therefore, $P = P'$. Whence, $CF \cap BE = D$. This implies that $\angle EPF = \angle BPC = 2\alpha$. According to the first lemma, $\angle ET'F = \angle EAF$ and $T'$ is on the perpendicular bisector of $EF$. Let $R$ be the intersection point of the common tangents of $QEF$ and $PEF$, taking into account that $R$ is the center of the homothety of these circles, it follows that $\angle ERF = \angle EPF - \angle EQF = \alpha$. Therefore, $R, O_1$, and $O_2$ would be collinear and $R$ is on the perpendicular bisector of $O_1O_2$. Hence, $R$ is the intersection of the corresponding arc of the $EF$ and its perpendicular bisector. Since such a point would be unique, we find that $R = T'$. We are done. ■