Iranian Mathematical Olympiad

First we start with two lemmas.

Lemma. Let $ABCD$ be a quadrilateral and $P$ be a point on segment $AB$. Let $X, Y, Z$ be points on $BC, CD, AD$ respectively such that $PX \parallel AC, XY \parallel BD, YZ \parallel AC$. Then we have $PZ \parallel BD$.

Proof. From $PX \parallel AC, XY \parallel BD, YZ \parallel AC$ we conclude that
$$\left. \begin{array}{l} \frac{BP}{AB} = \frac{BX}{BC} \\[2ex] \frac{BX}{BC} = \frac{DY}{DC} \\[2ex] \frac{DY}{DC} = \frac{DZ}{AD} \end{array} \right\} \implies \frac{BP}{AB} = \frac{DZ}{AD} \implies PZ \parallel BD$$

Lemma. Let $PQRS$ be a cyclic quadrilateral. $H$ is the intersection point of $PQ, RS$. Let $T'$ be an arbitrary point on $PQ$. $M, N$ are points on $PR, QS$ respectively such that $T'M \parallel QR, T'N \parallel PS$. $T$ is a point such that $\overline{T'MT} = \overline{T'NT} = 90^\circ$. Now that if $T'$ varies on line $PQ$ then $T$ varies on a line passing $H$ and if this line is $RS$ then $PS \perp QR$.

Proof. Suppose that there are three points $T'_1, T'_2, T'_3$ on $PQ$. Obviously we have
$$\frac{N_1 N_2}{N_1 N_3} = \frac{T'_1 T'_2}{T'_1 T'_3} = \frac{M_1 M_2}{M_1 M_3}$$
Let $A_1, A_2, A_3$ be the intersection points of $(T_1M_1, T_3N_3), (T_1M_1, T_2N_2), (T_2M_2, T_3N_3)$ respectively.
Because of intercept theorem we have
$$\frac{A_1 A_2}{A_1 T_1} = \frac{A_1 A_3}{A_1 T_3}, T_2 A_2 \parallel A_1 T_3, T_2 A_3 \parallel A_1 T_1$$
So we conclude that $T_2$ lies on $T_1T_3$.
We proved that $T$ varies on a line. Now we prove that this line passes through $H$.
Suppose that $SP, QR$ are not perpendicular. Then there exist $F, X', Y$. Set $T'_1 = Q, T'_2 = P$. So we have $T_1 = X', T_2 = Y$. Let $E, Z$ be the intersection points of $(PS, QR), (RX', SY)$ respectively. Obviously we have $EF \perp PQ, \overline{ZES} = 90^\circ - \overline{ERS}$. It means that
$$\overline{ZES} = 90^\circ - \overline{EPQ} \implies ZE \perp PQ$$
So $E, F, Z$ are collinear. From desargue theorem in triangles $SZR, PFQ$ we obtain that $PQ, RS, X'Y$ are concurrent. So $X'Y$ passes from the intersection point of $PQ, RS$.
Now suppose that $PQ \perp RS$. Then if we set $T'_1 = P, T'_2 = Q$ then we have $T_1 = S, T_2 = R$ which implies that $RS$ passes through $H$.

Case 1:
Suppose that $\widehat{PBA} + \widehat{QCA} = 90^\circ$. Let $M, N, T'$ be the second intersection points of $BP, CQ, AP$ with the circumcircle of triangle $ABC$. $PQ$ is the external angle bisector of $\hat{A}$. Thus we have
$$\begin{aligned} \widehat{QAC} = \widehat{PAB} &= 90^\circ - \frac{\hat{A}}{2} \implies \widehat{APX} = \widehat{AQX} = \frac{\hat{A}}{2} \\ \implies PX = QX, \widehat{PXQ} &= 180^\circ - \hat{A} \end{aligned}$$
We know that $PX = RX, QX = SX$ so we infer that
$$PX = QX = RX = SX \implies PQRS \text{ is cyclic}$$
$T'ACN$ is a cyclic quadrilateral. So $QNT' = 90^\circ - \frac{\hat{A}}{2}$.
$$\begin{aligned} PQRS \text{ is cyclic, } \widehat{PXQ} &= 180^\circ - \hat{A} \implies \widehat{QSP} = 90^\circ - \frac{\hat{A}}{2} \\ \implies \widehat{QNT'} = \widehat{QSP} \implies NT' \parallel PS \end{aligned}$$
By the same way we can see that $MT \parallel PS$.
$$\begin{aligned} \widehat{PBA} + \widehat{PCA} &= 90^\circ \\ \implies \widehat{XQC} + \widehat{XPB} &= 90^\circ \implies \widehat{PXB} + \widehat{QXC} = 180^\circ \\ \implies \widehat{RXS} &= \hat{A} \\ \implies \widehat{RQS} = \widehat{RPS} &= \frac{\hat{A}}{2} = \widehat{TNC} = \widehat{TMB} \implies TN \parallel QR, TM \parallel PS \end{aligned}$$

By the first lemma we conclude that the intersection point of $MT$, $NT$ lies on $RS$. So $R$, $S$, $T$ are collinear.

Case 2:
Suppose that $R$, $S$, $T$ are collinear. Likewise the first case we can see that $PQRS$ is cyclic with circumcenter $X$ and $T' M \parallel QR$, $T' N \parallel PS$. In circumcircle of triangle $ABC$, $TT'$ is diameter. So we have
$$\widehat{TMT'} = \widehat{TNT'} = 90^\circ$$
By applying the second lemma in cyclic quadrilateral $PQRS$ we conclude that $T$ lies on a line passing from the intersection point of $PQ, RS$. We know that $T$ lies on $RS$. So we have $PS \perp QR$ or $T$ is the intersection point of $PQ, RS$.
If $T$ lies on $PQ, RS$ we obtain that $T, T', A$ are collinear which is a contradiction. If $PS \perp QR$ it means that
$$\widehat{PXB} + \widehat{QXC} = 180^\circ \implies \widehat{XQC} + \widehat{XPB} = 90^\circ \implies \widehat{PBA} + \widehat{QCA} = 90^\circ$$