Problem 2251

Solution:

We work with directed angles modulo $180$ degrees. First we recognize that from the assumption $|AC| = |BC|$ it follows immediately that the angles $\angle BAC$, $\angle CBA$, $\angle DCA$ and $\angle ADC$ are all equal in size.

It suffices to show that the intersection point $S$ of the lines $AQ$ and $CD$ lies on the line $BR$. Since $\angle CSQ = \angle PAQ = \angle CRQ$, $CSRQ$ is a cyclic quadrilateral. Since $\angle CRA = 180^{\circ} - \angle ARP = 180^{\circ} - \angle AQP = \angle DQA = \angle DCA = \angle CBA$, $ABRC$ is also a cyclic quadrilateral. Hence $\angle SRC = \angle SQC = \angle ADC = \angle BAC = \angle BRP$ holds, from which it indeed follows that $P$, $R$ and $S$ lie on a line.

Solution:

As in the first solution, one shows that $ABRC$ is a cyclic quadrilateral. We consider the intersection point $E$ of the lines $AD$ and $BR$. Then $\angle EAP = \angle CBA = \angle BAC = \measuredangle ERP$ holds, so $E$ lies on the circumcircle of the cyclic quadrilateral $APRQ$.

Now $\angle CER = \angle APR = 180^{\circ} - \angle DCR$ holds, so $E$ also lies on the circumcircle of triangle $RCD$. The pairwise radical axes of the circumcircles of $ACD$, $APR$ and $RCD$ are thus the lines $CD$, $AQ$, $RE = RB$, so they meet at a single point. This implies the claim.

Solution:

Since $\angle QAC = \measuredangle QDC = \angle QPA$, it follows by the converse of the tangent-chord angle theorem that the line $AC$ is tangent to the circumcircle of the cyclic quadrilateral $APRQ$. By assumption, $AB$ is also tangent to the circumcircle of the cyclic quadrilateral $AQCD$, so the angles $\angle ADQ$ and $\angle PAQ$ are equal in size. From this it follows that $\angle ACQ = \angle ADQ = \angle PAQ = \angle CRQ$, so $AC$ is also tangent to the circumcircle of triangle $CQR$.

The intersection point $M$ of the lines $RQ$ and $AC$ thus satisfies, by the secant-tangent theorem, $|MA|^{2} = |MQ| \cdot |MR| = |MC|^{2}$, so $M$ is the midpoint of the segment $AC$, and hence also the midpoint of the segment $BD$.

Pappus's theorem, applied to the respectively collinear points $A, B, P$ and $R, Q, M$, now implies that the intersection point $S$ of the lines $AQ$ and $BR$, $C$ (as the intersection point of the lines $AM$ and $PR$), and $D$ (as the intersection point of the lines $BM$ and $PQ$) lie on a line, which is exactly the statement to be shown.