Subfield: Geometry


Question: Let $A B C$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $\mathrm{ABC}$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $E C$ and $F B$ respectively such that $\angle E M A=\angle B C A$ and $\angle A N F=\angle A B C$. Prove that the points $E, F, N, M$ lie on a circle.

Solution:
The first solution is based on the main Lemma. We present this Lemma with two different proofs.

Lemma: Let ABC be an acute triangle with AB=BC. Let P be any point on AC. Line passing through P perpendicular to AB, intersects ray BC in point T. If the line AT intersects the circumscribed circle of the triangle ABC the second time at point K, then $\angle AKP=\angle ABP$

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Proof 1:
Let $H$ be the orthocenter of the triangle $A B P$. Then

$$
\angle B H P=180^{\circ}-\angle B A C=180^{\circ}-\angle B C P .
$$

So $B H P C$ is cyclic. Then we get

$$
T K \cdot T A=T C \cdot T B=T P \cdot T H .
$$

So, $A H P K$ is also cyclic. But then

$$
\angle A K P=180^{\circ}-\angle A H P=\angle A B P .
$$



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Proof 2:
Consider the symmetric points $B^{\prime}$ and $C^{\prime}$ of $B$ and $C$, respectively, with respect to the line $P T$. It is clear that

$$
\mathrm{TC}^{\prime} \cdot T B^{\prime}=T C \cdot T B=T K \cdot T A
$$

So $B^{\prime} C^{\prime} K A$ is cyclic. Also, because of the symmetry we have

$$
\angle P C^{\prime} B^{\prime}=\angle P C B=\angle P A B .
$$

So $B^{\prime} C^{\prime} P A$ is also cyclic. Therefore, the points $B^{\prime}, C^{\prime}, K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have

$$
\angle P K A=\angle P B^{\prime} A=\angle P B A .
$$

So, lemma is proved and now return to the problem.



Let $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M^{\prime}$ and $N^{\prime}$ the intersection points of the circumcircle of the triangle $H E F$ with the segments $E C$ and $F B$, respectively. We are going to show that $M=M^{\prime}$ and $N=N^{\prime}$ and it will prove the points $E, F, N, M$ lie on a common circle.

Of course, $A$ is an orthocenter of the triangle $B C H$. Therefore $\angle B H A=\angle B C A, \angle C H A=\angle C B A$ and $\angle H B A=\angle H C A$. Thus

$$
\angle H E F=\angle H B A+\angle E A B=\angle H C A+\angle F A C=\angle H F E .
$$

So, the triangle $H E F$ is isosceles, $H E=H F$.

By using lemma, we get

$$
\angle A M^{\prime} E=\angle A H E=\angle A C B,
$$

and

$$
\angle A N^{\prime} F=\angle A H F=\angle A B C \text {. }
$$

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Therefore $M=M^{\prime}$ and $N=N^{\prime}$ and we are done.
Let $X, Y$ be projections of $B$ on $A C$, and $C$ on $A B$, respectively. Let $\omega$ be circumcircle of $B X Y C$. Let $Z$ be intersection of $E C$ and $\omega$ and $D$ be projection of $E$ on $B A$.

$$
\angle M A C=\angle A M E-\angle M C A=\angle X C B-\angle X C E=\angle Z C B=\angle Z X B
$$

Since $B X Y C$ is cyclic $\angle A C Y=\angle X B A$, and since DEXA is cyclic

$$
\angle E X D=\angle E A D=\angle F A C \text {. }
$$

Therefore, we get that the quadrangles $B Z X D$ and $C M A F$ are similar. Hence $\angle F M C=\angle D Z B$. Since $Z E D B$ is cyclic,

$$
\angle D Z B=\angle D E B=\angle X A B .
$$

Thus $\angle F M C=\angle X A B$. Similarly, $\angle E N B=\angle Y A C$. We get that $\angle F M C=\angle E N B$ and it implies that the points $E, F, N, M$ lie on a circle.

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