Subfield: Geometry


Question: Let $A B C D$ be a parallelogram such that $A C=B C$. A point $P$ is chosen on the extension of the segment $A B$ beyond $B$. The circumcircle of the triangle $A C D$ meets the segment $P D$ again at $Q$, and the circumcircle of the triangle $A P Q$ meets the segment $P C$ again at $R$. Prove that the lines $C D, A Q$, and $B R$ are concurrent.

Solution:
Since $A C=B C=A D$, we have $\angle A B C=\angle B A C=\angle A C D=\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain

$$
\angle C R A=180^{\circ}-\angle A R P=180^{\circ}-\angle A Q P=\angle D Q A=\angle D C A=\angle C B A,
$$

so the points $A, B, C$, and $R$ lie on some circle $\gamma$.


Introduce the point $X=A Q \cap C D$; we need to prove that $B, R$ and $X$ are collinear.

By means of the circle $(A P R Q)$ we have

$$
\angle R Q X=180^{\circ}-\angle A Q R=\angle R P A=\angle R C X
$$

(the last equality holds in view of $A B \| C D$ ), which means that the points $C, Q, R$, and $X$ also lie on some circle $\delta$.

Using the circles $\delta$ and $\gamma$ we finally obtain

$$
\angle X R C=\angle X Q C=180^{\circ}-\angle C Q A=\angle A D C=\angle B A C=180^{\circ}-\angle C R B,
$$

that proves the desired collinearity.

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Since $A C=B C=A D$, we have $\angle A B C=\angle B A C=\angle A C D=\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain

$$
\angle C R A=180^{\circ}-\angle A R P=180^{\circ}-\angle A Q P=\angle D Q A=\angle D C A=\angle C B A,
$$

so the points $A, B, C$, and $R$ lie on some circle $\gamma$.



Let $\alpha$ denote the circle $(A P R Q)$. Since

$$
\angle C A P=\angle A C D=\angle A Q D=180^{\circ}-\angle A Q P
$$

the line $A C$ is tangent to $\alpha$.

Now, let $A D$ meet $\alpha$ again at a point $Y$ (which necessarily lies on the extension of $D A$ beyond $A$ ). Using the circle $\gamma$, along with the fact that $A C$ is tangent to $\alpha$, we have

$$
\angle A R Y=\angle C A D=\angle A C B=\angle A R B
$$

so the points $Y, B$, and $R$ are collinear.

Applying Pascal's theorem to the hexagon $A A Y R P Q$ (where $A A$ is regarded as the tangent to $\alpha$ at $A$ ), we see that the points $A A \cap R P=C, A Y \cap P Q=D$, and $Y R \cap Q A$ are collinear. Hence the lines $C D, A Q$, and $B R$ are concurrent.
Since $A C=B C=A D$, we have $\angle A B C=\angle B A C=\angle A C D=\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain

$$
\angle C R A=180^{\circ}-\angle A R P=180^{\circ}-\angle A Q P=\angle D Q A=\angle D C A=\angle C B A,
$$

so the points $A, B, C$, and $R$ lie on some circle $\gamma$.



we introduce the point $X=A Q \cap C D$ and aim at proving that the points $B, R$, and $X$ are collinear. we denote $\alpha=(A P Q R)$; but now we define $Y$ to be the second meeting point of $R B$ with $\alpha$.

Using the circle $\alpha$ and noticing that $C D$ is tangent to $\gamma$, we obtain

$$
\angle R Y A=\angle R P A=\angle R C X=\angle R B C .
$$

So $A Y \| B C$, and hence $Y$ lies on $D A$.

Now the chain of equalities (1) shows also that $\angle R Y D=\angle R C X$, which implies that the points $C, D, Y$, and $R$ lie on some circle $\beta$. Hence, the lines $C D, A Q$, and $Y B R$ are the pairwise radical axes of the circles $(A Q C D), \alpha$, and $\beta$, so those lines are concurrent.