Subfield: Geometry


Question: Let $A B C$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $B C, C A, A B$ respectively. One of the intersection points of the line $E F$ and the circumcircle is $P$. The lines $B P$ and $D F$ meet at point $Q$. Prove that $A P=A Q$.

Solution:
The line $E F$ intersects the circumcircle at two points. Depending on the choice of $P$, there are two different cases to consider.

Case 1: The point $P$ lies on the ray $E F$ (see Fig. 1).

Let $\angle C A B=\alpha, \angle A B C=\beta$ and $\angle B C A=\gamma$. The quadrilaterals $B C E F$ and $C A F D$ are cyclic due to the right angles at $D, E$ and $F$. So,

$$
\begin{aligned}
& \angle B D F=180^{\circ}-\angle F D C=\angle C A F=\alpha, \\
& \angle A F E=180^{\circ}-\angle E F B=\angle B C E=\gamma, \\
& \angle D F B=180^{\circ}-\angle A F D=\angle D C A=\gamma .
\end{aligned}
$$

Since $P$ lies on the arc $A B$ of the circumcircle, $\angle P B A<\angle B C A=\gamma$. Hence, we have

$$
\angle P B D+\angle B D F=\angle P B A+\angle A B D+\angle B D F<\gamma+\beta+\alpha=180^{\circ}
$$

and the point $Q$ must lie on the extensions of $B P$ and $D F$ beyond the points $P$ and $F$, respectively.

From the cyclic quadrilateral $A P B C$ we get

$$
\angle Q P A=180^{\circ}-\angle A P B=\angle B C A=\gamma=\angle D F B=\angle Q F A .
$$

Hence, the quadrilateral $A Q P F$ is cyclic. Then $\angle A Q P=180^{\circ}-\angle P F A=\angle A F E=\gamma$.

We obtained that $\angle A Q P=\angle Q P A=\gamma$, so the triangle $A Q P$ is isosceles, $A P=A Q$.

<img_3820>

Fig. 1

<img_3703>

Fig. 2



Case 2: The point $P$ lies on the ray $F E$ (see Fig. 2). In this case the point $Q$ lies inside the segment $F D$.

Similarly to the first case, we have

$$
\angle Q P A=\angle B C A=\gamma=\angle D F B=180^{\circ}-\angle A F Q .
$$

Hence, the quadrilateral $A F Q P$ is cyclic.

Then $\angle A Q P=\angle A F P=\angle A F E=\gamma=\angle Q P A$. The triangle $A Q P$ is isosceles again, $\angle A Q P=\angle Q P A$ and thus $A P=A Q$.
For arbitrary points $X, Y$ on the circumcircle, denote by $\overparen{X Y}$ the central angle of the $\operatorname{arc} X Y$.

Let $P$ and $P^{\prime}$ be the two points where the line $E F$ meets the circumcircle; let $P$ lie on the $\operatorname{arc} A B$ and let $P^{\prime}$ lie on the $\operatorname{arc} C A$. Let $B P$ and $B P^{\prime}$ meet the line $D F$ and $Q$ and $Q^{\prime}$, respectively (see Fig. 3). We will prove that $A P=A P^{\prime}=A Q=A Q^{\prime}$.

<img_3435>

Fig. 3

Like in the first solution, we have $\angle A F E=\angle B F P=\angle D F B=\angle B C A=\gamma$ from the cyclic quadrilaterals $B C E F$ and $C A F D$.

By $\overparen{P B}+\overparen{P^{\prime} A}=2 \angle A F P^{\prime}=2 \gamma=2 \angle B C A=\overparen{A P}+\overparen{P B}$, we have

$$
\overparen{A P}=\overparen{P^{\prime} A}, \quad \angle P B A=\angle A B P^{\prime} \quad \text { and } \quad A P=A P^{\prime} \text {. }\tag{1}
$$

Due to $\overparen{A P}=\overparen{P^{\prime} A}$, the lines $B P$ and $B Q^{\prime}$ are symmetrical about line $A B$.

Similarly, by $\angle B F P=\angle Q^{\prime} F B$, the lines $F P$ and $F Q^{\prime}$ are symmetrical about $A B$. It follows that also the points $P$ and $P^{\prime}$ are symmetrical to $Q^{\prime}$ and $Q$, respectively. Therefore,

$$
A P=A Q^{\prime} \quad \text { and } \quad A P^{\prime}=A Q .\tag{2}
$$

The relations (1) and (2) together prove $A P=A P^{\prime}=A Q=A Q^{\prime}$.