I have a solution but it uses some advanced Euclidean geometry stuff: inversion with respect to a circle and the polar relation. Call k the circle with diameter EF. Let the line PQ meet the line BC at T and the line AM at S. Finally, let H be the foot of the A-altitude of triangle ABC. Since the points P and Q lie on the circle with diameter EF, the center M of this circle is equidistant from the lines CA and AB. Hence, this center lies on the angle bisector of the angle CAB. Thus, < MAP = < MAQ; also, we have < APM = < AQM = 90 and AM = AM. Thus, the triangles MAP and MAQ are congruent; for this reason, the points P and Q are symmetric with respect to the line AM, and consequently, we have AP = AQ, and the triangle PAQ is isosceles. The line AM is the axis of symmetry of this triangle (since < MAP = < MAQ); hence, AM is perpendicular to PQ. It follows that the triangle MSQ is right-angled. But we know that the triangle MQA is right-angled, too. These two right-angled triangles are similar (since < QMS = < AMQ); hence, MS / MQ = MQ / MA, so that $MS\cdot MA=MQ^{2}$. But M is the center and MQ is the radius of the circle k. Therefore, the point S is the image of the point A in the inversion with respect to k. The line PQ is therefore the perpendicular to the line AM through the image of A in the inversion with respect to k. In other words, the line PQ is the polar of A with respect to k. The point T lies on the line PQ; hence, the point T lies on the polar of A with respect to k. Therefore, by a well-known theorem, conversely, the point A lies on the polar of T with respect to k. Now, by definition, the polar of T with respect to k must be perpendicular to the line TM, i. e. to the line BC. Hence, the polar of T with respect to k is the perpendicular to BC through A, i. e. it is the altitude of triangle ABC from the vertex A. Now, we want to show that the point of intersection of the lines EP and FQ lies on this altitude. Well, there is a well-known theorem stating that if a quadrilateral ABCD is inscribed in a circle k, then the point of intersection of the sidelines AB and CD lies on the polar of the point of intersection of the sidelines BC and DA with respect to k. Applying this to the quadrilateral EPQF, we find that the point of intersection of the sidelines EP and QF lies on the polar of the point of intersection of the sidelines PQ and FE with respect to k. But the point of intersection of PQ and FE is T, and its polar with respect to k is the altitude from A. Hence, we conclude that the point of intersection of EP and QF lies on the altitude from A. Proof complete. I'm sorry, but I have no idea how to find a rather elementary proof. Hopefully some others can help. Darij

If you consider $A'$ intersection of $EQ$ and $FP$ and $A''$ intersection of $EP$ and $FQ$ then $A'A''$is perpendicular to $BC$. And $A$ is midpoint of $A'A''$

I didn't understand your solution Darij because I don't know what "inversion and polar with respect to..." are.I guess they are geometrical transformations,but i've never heard about them... Could you explain them to me,please?At least only the definitions,so that i can understand your solution. Maybe it's easy,but how do you prove that A'A" is perpendicular to BC and A is the midpoint,Omid?Could you send the full solution please? Thanks to both of you.

Omid, thanks, a very nice proof! Iris, you asked why A'A" is perpendicular to BC. In fact, the three altitudes of a triangle always concur. Now, take the triangle A'EF. Two altitudes of this triangle are EP and FQ (this is because < EPF = 90 and < EQF = 90; remember P and Q lie on the circle with diameter EF). Hence, the third altitude passes through the point of intersection of these two altitudes. In other words, the perpendicular from A' to BC passes through the point of intersection of the altitudes EP and FQ, i. e. through the point A". Hence, A'A" is perpendicular to BC. You also asked why A is the midpoint of A'A". This is a bit tricky. Since < A'PA" = 90 and < A'QA" = 90, the points P and Q lie on the circle with diameter A'A". The center S of this circle must be the midpoint of A'A". Now, we want to prove that S = A. We have < PSQ = 2 < PA'Q, in other words < PSQ = 2 < FA'Q. But since SP = SQ, the triangle PSQ is isosceles, so that < SPQ = (180 - < PSQ) / 2 = (180 - 2 < FA'Q) / 2 = 90 - < FA'Q = < QFA' = < QFP = < QMP / 2 (since M is the midpoint of the circle with diameter EF). But the quadrilateral APMQ yields < QMP = 360 - < MPA - < PAQ - < AQM = 360 - 90 - A - 90 = 180 - A, so that < SPQ = < QMP / 2 = (180 - A) / 2. On the other hand, in the isosceles triangle PAQ, we have < APQ = (180 - < PAQ) / 2 = (180 - A) / 2. So that < SPQ = < APQ; hence, the point S lies on the line AP. Similarly, the point S lies on the line AQ. But now there is nothing left for the point S other than to coincide with A ! So that S = A. Now, it follows that A is the midpoint of A'A". [Omid, sorry, I have a tendency to make things more complicated than they actually are; I guess there is a much simpler proof here.] Iris Aliaj wrote: I didn't understand your solution Darij because I don't know what "inversion and polar with respect to..." are.I guess they are geometrical transformations, Indeed. Inversion is an important weapon in geometry (whole books are written about this transformation), but I need nothing but its definition! If a circle k has the center M and the radius r, and P is an arbitrary point in the plane (not coinciding with M), then let P' be the point on the half-line MP such that $MP\cdot MP^{\prime }=r^{2}$. Then, this point P' is called the image of the point P in the inversion with respect to the circle k. Of course, since the equation $MP\cdot MP^{\prime }=r^{2}$ is symmetric in P and P', the point P' is also the image of P in the inversion with respect to the circle k. Instead of "image of P in the inversion with respect to the circle k", one usually uses the shorter paraphrase "inverse of P in the circle k". If P = M, then the inverse of P in k is, strictly speaking, not defined, but one uses to say that the inverse of P in k is the "point at infinity". Now, the inversion with respect to the circle k is the geometrical transformation mapping each point P in the plane to the inverse of P in the circle k. The polar relationship is a rather unlikely kind of transformation: It maps points to lines and lines to points! In fact, let again P be a point in the plane of a circle k, but not coinciding with the center M of k. Now, let P' be the inverse of P in k. Then, consider the perpendicular g to the line MP through the point P'. This line g is called the polar of the point P with respect to the circle k. Hence, the notion of "polar" attributes to each point a line, namely the polar of the point with respect to a circle. Now, there is another notion, attributing to each line a point: Let g be a line in the plane of the circle k not passing through the center M of k. Then, let the perpendicular from M to g meet g at T, and let T' be the inverse of T in k. Then, the point T' is called the pole of the line g with respect to the circle k. Of course, you easily see that if g is the polar of a point P with respect to a circle k, then P is the pole of the line g with respect to k. In other words, the transformation "line --> its pole" is the converse of the transformation "point --> its polar" (of course, both times with respect to one given circle). Note that if a line g passes through the center M of the circle k, then the pole of g with respect to k is undefined again, but again one uses to say that it is the "infinite point in the direction perpendicular to g". But again we won't need this in the further. In my above solution of your problem, I used two important theorems: Theorem 1. If k is a circle, and the point P lies on the polar of another point Q with respect to k, then the point Q lies on the polar of P with respect to k. Theorem 2. If k is a circle, and ABCD is a quadrilateral inscribed in k, then the point of intersection of the sidelines AB and CD lies on the polar of the point of intersection of the sidelines BC and DA with respect to k. Theorem 1 is quite simple, while Theorem 2 is rather difficult. Darij

Why is PAQ necesarily isosceles? We are trying to prove that A=S, so we cannot assume that, no? if it weere isosceles, then of course A is on A'A''.

Assume that $K \equiv EP \cap FQ$ and $D\equiv AK \cap BC$. $\triangle QAP$ is isosceles, thus $\angle AQP = \frac{\angle B +\angle C}{2}$. Also $\angle QEK = \angle AQP$. Thus, in $\triangle QEK$, $\angle QKE = 90 - \angle QEK = \frac {\angle A}{2}$. Note that in $AQKP$, $\angle QKP = 180 - \frac{\angle A}{2}$. Thus, A is the circumcentre of $\triangle QKP$ and thus $AK = AP$. Thus, $\angle AKP = \angle APK = \angle PFE$, which implies $KPFD$ is cyclic and hence $AD$ perpendicular to $BC$.

Example 1 in the link below pretty much explains it: http://www.math.ust.hk/excalibur/v11_n4.pdf

Omid Hatami wrote: If you consider $A'$ intersection of $EQ$ and $FP$ and $A''$ intersection of $EP$ and $FQ$ then $A'A''$is perpendicular to $BC$. And $A$ is midpoint of $A'A''$ Can someone explain why $A$ the midpoint of $A'A''$?

Pardon the mislabeled points :/

Let $ PQ $ meet $ BC $ at $ D $. Then the polar of $ A $ wrt the circle passes through $ D $. So the polar of $ D $ passes through $ A $. But we know that it passes through $ K $ too. So the polar of $ D $ is $ AK $ which is therefore perpendicular to $ BC $.

Omid Hatami wrote: If you consider $A'$ intersection of $EQ$ and $FP$ and $A''$ intersection of $EP$ and $FQ$ then $A'A''$is perpendicular to $BC$. And $A$ is midpoint of $A'A''$ Even if we prove that $A',A',A''$ points are collinear then we complete proof. .

Let $ D $ such that $ DK $ is perpendicular to $ BC $. Draw line $ AK $, now we want to prove that angle of $ AKP $+ angle of $ PKF $+ angle of $ FKD $ is 180. let angle of $ QFE $ is b, angle of $ FPC $ is x, angle of $ PEF $ is a, and because $ EF $ is diameter, so angle of $ EPF $= angle of $ EQF $= 90, with angle chasing, we will have angle of $ AKD $ is 180

Let $J$ be the intersection of lines $EQ$ and $FP$. Using Pascal's theorem in the hexagon $FPPEQQ$ we get $J,A,K$ collinear. $K$ is the orthocenter of $\triangle ABC$ thus $JK \perp BC$ $=>$ $AK \perp BC$ $\blacksquare$ (this is suprisingly trivial even for a junior olympiad)

We begin by showing that $A$ is the circumcenter of $\triangle KPQ$: Consider the configuration where $\angle ACB$ is obtuse and diameter $EF$ lies on extended line $BC$. Let $O$ be the midpoint of diameter $EF$. Let $KA$ meet $BC$ at $G$. Let us define $\angle KPA = \alpha$ and $\angle KQA = \beta$ By applying Tangent Chord Angle theorem, we get: $\angle POE = 2\alpha$ and $\angle QOF = 2\beta$ Now, $\angle POQ = 180 - 2(\alpha + \beta)$, and since $PAQO$ is a cyclic quadrilateral, we have $\angle PAQ = 2(\alpha + \beta)$ Now $\angle POF = 180 - 2\alpha$, so $\angle PEF = 90 - \alpha$ Similarly, we have $\angle QOE = 180 - 2\beta$, so $\angle QFE = 90 - \beta$ From $\triangle EKF, \angle EKF = 180 - (\angle PEF + \angle QFE)$ $= 180 - (90 - \alpha + 90 - \beta) = (\alpha + \beta)$ Thus, we have $\angle PKQ = \angle EKF = \angle PAQ/2$ Also, $AP = AQ$ (Since $AP$ and $AQ$ are tangents to the same circle) From the above 2 results, it readily follows that $A$ is the circumcenter of $\triangle KPQ$. Thus, we have $AK = AP$, and so $\angle AKP = \angle KPA = \alpha$ So in $\triangle EKG, \angle KGE = 180 - (\angle AKP + \angle PEF)$ $= 180 - (\alpha + 90 - \alpha) = 90^{\circ}$ So $KA$ is perpendicular to $BC$, hence $K$ lies on the altitude from $A$ of the triangle $ABC$.

Let $X = FP\cap QE$. Since $\angle EPF = \angle FQE = 90^\circ$, $K$ is the orthocenter of $\triangle XEF$, so $XK\perp EF$. Hence, it suffices to show that $A$ lies on line $XK$. But this follows directly from Pascal's Theorem on $PPFQQE$, so we are done.

1st problem of the China national olympiad 1996 https://artofproblemsolving.com/community/c6h359386p1964606

a Pascal solution like #16 , writing for the joy having found it on my own