Problem 1561

Fig. 42
Let us consider the case where the ray $AB$ is a bisector of the angle $PAQ$, and $\angle BAC < \angle BAQ$ (fig. 42). We will prove that in this case the ray $CD$ is the bisector of the angle $PCQ$. In the case where $AD$ is the bisector of the angle $PAQ$ all the proof is analogous.

Let $M$ be the midpoint of $PQ$, and $O$ the intersection point of the diagonals of the rectangle. Let us select point $S$ symmetrically to $Q$ with respect to the point $O$. In this case $AQCS$ is a parallelogram.

Let us first prove that $\angle APC = \angle AQC$. For this we should prove that the points $A$, $S$, $P$, $C$ are cyclic. Indeed, let $\angle BAC = \alpha$, and $\angle BAQ = \beta$. By the assumption, $\beta > \alpha$. Then $\angle CAQ = \beta - \alpha$.

Also $CS \parallel AQ$, then $\angle SCA = \angle CAQ = \beta - \alpha$. $AB$ is the bisector of the angle $PAQ$, therefore $\angle PAB = \beta$. Moreover, $ABCD$ is a rectangle, which leads to $\angle ABD = \angle BAC = \alpha$. $M$ is the midpoint of $PQ$, and $O$ is the midpoint of $QS$, hence $MO$ is the mid-segment of $PQS$. This brings the fact that $PS \parallel MO$, i.e. $PS \parallel BD$. Which leads to $\angle PTB = \angle TBD = \alpha$, where $T = (PS) \cap (AB)$.

According to the Triangle exterior angle theorem, we obtain that
$$\angle APS = \angle APT = \angle PAB - \angle PTA = \beta - \alpha.$$
Thus, we proved that $\angle APS = \angle ACS$, i.e. the points $A$, $S$, $P$, $C$ are cyclic. From which it follows that $\angle APC = \angle ASC = \angle AQC$ (because inscribed angles that have the same arc are of the same length, and also the opposite angles of a parallelogram are the same).

Fig. 43
What is left is to prove that when in the convex quadrangle $APCQ$ the opposite angles $APC$ and $AQC$ are equal, then the bisectors of the angles $PAQ$ and $PCQ$ are parallel.

Let the bisector of the angle $PAQ$ intersect lines $PC$ and $CQ$ in the points $N$ and $F$ respectively (fig. 43), and let the bisector of the angle $PCQ$ intersect $AQ$ in the point $L$. Let $K$ be the point symmetrical to $P$ with respect to $AF$, then the point $K$ lies on $AQ$ (as $AF$ is the bisector of the angle $PAQ$). This means that $\angle AKN = \angle APN$. $\angle APC = \angle AQC$, therefore $\angle AKN = \angle AQC$. Thus, $NK \parallel CQ$. Which means $\angle CFN = \angle KNA = \angle PNA = \angle CNF$, i.e. $\angle CFN = \angle CNF$ and triangle $NCF$ is isosceles. The bisector of the exterior angle adjacent to the vertex of isosceles triangle is parallel to its base, therefore $AF \parallel CL$, Q.F.D.

In the case when $\alpha > \beta$, the points $P$ and $Q$ will lie in one direction from the line $AC$ and it will hold that $\angle APC + \angle AQC = 180^\circ$. The ray $CB$ will be the bisector of the angle $PCQ$ (all proofs are analogous).