Subfield: Geometry


Question: Let $A B C$ be a triangle and let $D$ be a point on the segment $B C, D \neq B$ and $D \neq C$. The circle $A B D$ meets the segment $A C$ again at an interior point $E$. The circle $A C D$ meets the segment $A B$ again at an interior point $F$. Let $A^{\prime}$ be the reflection of $A$ in the line $B C$. The lines $A^{\prime} C$ and $D E$ meet at $P$, and the lines $A^{\prime} B$ and $D F$ meet at $Q$. Prove that the lines $A D, B P$ and $C Q$ are concurrent (or all parallel).

Solution:
Let $\sigma$ denote reflection in the line $B C$. Since $\angle B D F=\angle B A C=$ $\angle C D E$, by concyclicity, the lines $D E$ and $D F$ are images of one another under $\sigma$, so the lines $A C$ and $D F$ meet at $P^{\prime}=\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\prime}=\sigma(Q)$. Consequently, the lines $P Q$ and $P^{\prime} Q^{\prime}=\sigma(P Q)$ meet at some (possibly ideal) point $R$ on the line $B C$.



Since the pairs of lines $(C A, Q D),(A B, D P),(B C, P Q)$ meet at three collinear points, namely $P^{\prime}, Q^{\prime}, R$ respectively, the triangles $A B C$ and $D P Q$ are perspective, i.e., the lines $A D, B P, C Q$ are concurrent, by the Desargues theorem.



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Fig. 1



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Fig. 2
As in the first solution, $\sigma$ denotes reflection in the line $B C$, the lines $D E$ and $D F$ are images of one another under $\sigma$, the lines $A C$ and $D F$ meet at $P^{\prime}=\sigma(P)$, and the lines $A B$ and $D E$ meet at $Q^{\prime}=\sigma(Q)$.



Let the line $A D$ meet the circle $A B C$ again at $M$. Letting $M^{\prime}=\sigma(M)$, it is sufficient to prove that the lines $D M^{\prime}=\sigma(A D), B P^{\prime}=\sigma(B P)$ and $C Q^{\prime}=\sigma(C Q)$ are concurrent.







Begin by noticing that $\angle\left(B M^{\prime}, M^{\prime} D\right)=-\angle(B M, M A)=-\angle(B C, C A)=\angle(B F, F D)$, to infer that $M^{\prime}$ lies on the circle $B D F$. Similarly, $M^{\prime}$ lies on the circle $C D E$, so the line $D M^{\prime}$ is the radical axis of the circles $B D F$ and $C D E$.



Since $P^{\prime}$ lies on the lines $A C$ and $D F$, it is the radical centre of the circles $A B C, A D C$, and $B D F$; hence the line $B P^{\prime}$ is the radical axis of the circles $B D F$ and $A B C$. Similarly, the line $C Q^{\prime}$ is the radical axis of the circles $C D E$ and $A B C$. So the conclusion follows: the lines $D M^{\prime}$, $B P^{\prime}$ and $C Q^{\prime}$ are concurrent at the radical centre of the circles $A B C, B D F$ and $C D E$; thus the lines $D M, B P^{\prime}$ and $C Q^{\prime}$ are also concurrent.
As in the previous solutions, $\sigma$ denotes reflection in the line $B C$. Let the lines $B E$ and $C F$ meet at $X$. Due to the circles $B D E A$ and $C D F A$, we have $\angle X B D=$ $\angle E A D=\angle X F D$, so the quadrilateral $B F X D$ is cyclic; similarly, the quadrilateral $C E X D$ is cyclic. Hence $\angle X D B=\angle C F A=\angle C D A$, the lines $D X$ and $D A$ are therefore images of one another under $\sigma$, and $X^{\prime}=\sigma(X)$ lies on the line $A D$. Let $E^{\prime}=\sigma(E)$ and $F^{\prime}=\sigma(F)$, and apply the Pappus theorem to the hexagon $B P F^{\prime} C Q E^{\prime}$ to infer that $X^{\prime}, D$, and $B P \cap C Q$ are collinear. The conclusion follows.



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Fig. 3