Subfield: Geometry


Question: In an acute triangle $A B C$ segments $B E$ and $C F$ are altitudes. Two circles passing through the points $A$ and $F$ are tangent to the line $B C$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $P E$ and $Q F$ intersect on the circumcircle of triangle $A E F$.

Solution:
To approach the desired result we need some information about the slopes of the lines $P E$ and $Q F$; this information is provided by formulas (1) and (2) which we derive below.

The tangents $B P$ and $B Q$ to the two circles passing through $A$ and $F$ are equal, as $B P^{2}=B A \cdot B F=B Q^{2}$. Consider the altitude $A D$ of triangle $A B C$ and its orthocentre $H$. From the cyclic quadrilaterals $C D F A$ and $C D H E$ we get $B A \cdot B F=B C \cdot B D=B E \cdot B H$. Thus $B P^{2}=B E \cdot B H$, or $B P / B H=B E / B P$, implying that the triangles $B P H$ and $B E P$ are similar. Hence

$$
\angle B P E=\angle B H P \text {. }\tag{1}
$$

The point $P$ lies between $D$ and $C$; this follows from the equality $B P^{2}=B C \cdot B D$. In view of this equality, and because $B P=B Q$,

$$
D P \cdot D Q=(B P-B D) \cdot(B P+B D)=B P^{2}-B D^{2}=B D \cdot(B C-B D)=B D \cdot D C
$$

Also $A D \cdot D H=B D \cdot D C$, as is seen from the similar triangles $B D H$ and $A D C$. Combining these equalities we obtain $A D \cdot D H=D P \cdot D Q$. Therefore $D H / D P=D Q / D A$, showing that the triangles $H D P$ and $Q D A$ are similar. Hence $\angle H P D=\angle Q A D$, which can be rewritten as $\angle B P H=\angle B A D+\angle B A Q$. And since $B Q$ is tangent to the circumcircle of triangle $F A Q$,

$$
\angle B Q F=\angle B A Q=\angle B P H-\angle B A D .\tag{2}
$$

From (1) and (2) we deduce

$$
\begin{aligned}
& \angle B P E+\angle B Q F=(\angle B H P+\angle B P H)-\angle B A D=\left(180^{\circ}-\angle P B H\right)-\angle B A D \\
& =\left(90^{\circ}+\angle B C A\right)-\left(90^{\circ}-\angle A B C\right)=\angle B C A+\angle A B C=180^{\circ}-\angle C A B .
\end{aligned}
$$

Thus $\angle B P E+\angle B Q F<180^{\circ}$, which means that the rays $P E$ and $Q F$ meet. Let $S$ be the point of intersection. Then $\angle P S Q=180^{\circ}-(\angle B P E+\angle B Q F)=\angle C A B=\angle E A F$.

If $S$ lies between $P$ and $E$ then $\angle P S Q=180^{\circ}-\angle E S F$; and if $E$ lies between $P$ and $S$ then $\angle P S Q=\angle E S F$. In either case the equality $\angle P S Q=\angle E A F$ which we have obtained means that $S$ lies on the circumcircle of triangle $A E F$.

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Let $H$ be the orthocentre of triangle $A B C$ and let $\omega$ be the circle with diameter $A H$, passing through $E$ and $F$. Introduce the points of intersection of $\omega$ with the following lines emanating from $P: P A \cap \omega=\{A, U\}, P H \cap \omega=\{H, V\}, P E \cap \omega=\{E, S\}$. The altitudes of triangle $A H P$ are contained in the lines $A V, H U, B C$, meeting at its orthocentre $Q^{\prime}$.

By Pascal's theorem applied to the (tied) hexagon $A E S F H V$, the points $A E \cap F H=C$, $E S \cap H V=P$ and $S F \cap V A$ are collinear, so $F S$ passes through $Q^{\prime}$.

Denote by $\omega_{1}$ and $\omega_{2}$ the circles with diameters $B C$ and $P Q^{\prime}$, respectively. Let $D$ be the foot of the altitude from $A$ in triangle $A B C$. Suppose that $A D$ meets the circles $\omega_{1}$ and $\omega_{2}$ at the respective points $K$ and $L$.

Since $H$ is the orthocentre of $A B C$, the triangles $B D H$ and $A D C$ are similar, and so $D A \cdot D H=D B \cdot D C=D K^{2}$; the last equality holds because $B K C$ is a right triangle. Since $H$ is the orthocentre also in triangle $A Q^{\prime} P$, we analogously have $D L^{2}=D A \cdot D H$. Therefore $D K=D L$ and $K=L$.

Also, $B D \cdot B C=B A \cdot B F$, from the similar triangles $A B D, C B F$. In the right triangle $B K C$ we have $B K^{2}=B D \cdot B C$. Hence, and because $B A \cdot B F=B P^{2}=B Q^{2}$ (by the definition of $P$ and $Q$ in the problem statement), we obtain $B K=B P=B Q$. It follows that $B$ is the centre of $\omega_{2}$ and hence $Q^{\prime}=Q$. So the lines $P E$ and $Q F$ meet at the point $S$ lying on the circumcircle of triangle $A E F$.

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