Bulgarian Mathematical Competitions

Solution:

The Sine theorem for the triangles $ODQ$ and $AOQ$ gives
$$\frac{\sin \varphi}{\sin \beta} = \frac{QD}{OD}$$
and
$$\text{( + AOD) = AQ OA}$$
whence
$$\text{( + AOD) = AQ OA OD QD}$$

We obtain in the same way that
$$\text{( + AOB) = AP OA OB BP}$$
Therefore
$$\text{( + AOD) ( + AOB) = AQ AP OD OB BP QD}$$
We get in the same way that
$$\text{( DOC - ) ( BOC - ) = QC PC OB OD PD QB}$$
Using the Menelaus theorem for $\triangle ADC$ and the line $QP$ and for $\triangle ABC$ and the line $QP$ we obtain
$$\frac{AQ \cdot DP}{DQ \cdot CP} = \frac{AL}{CL} \text{ and } \frac{QC \cdot BP}{QB \cdot AP} = \frac{CL}{AL}$$
Setting $\text{+ AOD = x}$, $\text{+ AOB = y}$, $\text{DOC - = z}$ and $\text{BOC - = t}$, we have
$$\frac{\sin x \cdot \sin z}{\sin y \cdot \sin t} = \frac{AQ \cdot DP \cdot QC \cdot BP}{DQ \cdot CP \cdot QB \cdot AP} = \frac{AL}{CL} \cdot \frac{CL}{AL} = 1$$
i.e. $\sin x \cdot \sin z = \sin y \cdot \sin t$. It follows easily from here that
$$\cos (x-z) - \cos (x+z) = \cos (y-t) - \cos (y+t)$$
Since $x + y + z + t = 360^\circ$, we have $\cos (x+z) = \cos (y+t)$ and therefore $\cos (x-z) = \cos (y-t)$. Since $x-z + y-t < 360^\circ$ and the equality $x-z = t-y$ implies that $O$ lies on $PQ$ (prove this!) we obtain $x-z = y-t$, whence $x+t = z+y = 180^\circ$.