By construction $ABCP$ is cyclic so that $\angle APB=\angle BCA$. On the other hand since $AD$ and $CF$ are altitudes $ACDF$ is also cyclic, whence $\angle BFD=\angle BCA$. Putting this together one gets that $\angle APQ=\angle BFQ$, that is, $APQF$ is cyclic. Now by angle chasing $\angle AFE = \angle BFD = \angle BCA$, and using the previous fact gives $\angle AQP = \angle AFP =\angle BCA$. Thus $APQ$ is isosceles with $AP=AQ$. This was my argument for the case when $Q$ lies inside the triangle. When it is outside you do basically the same.

Sorry to double post, I just realized the cases can be avoided by using directed angles from the beginning: $\angle AFQ = \angle AFD = \angle ACD = \angle ACB \pmod \pi$ and $\angle APQ=\angle APB = \angle ACB \pmod \pi$, so $\angle APQ = \angle AFQ \pmod \pi$. But $\angle AQP = \angle AFP = \angle EFB= \angle ECB = \angle ACB \pmod \pi$. Therefore $\angle APQ = \angle AQP \pmod \pi$.

Hmm, this looks familiar...

$\angle{C}=\angle{APB}=\angle{BFC}=a=\angle{AEC}=>APQF$ is cyclic. $\angle{QAP}=\angle{DFP}=180-2a=>\angle{AQP}=a=>AP=AQ$

Let's correct the above, since it is indeed a short proof (although basically the proof at #2). $\angle{C}=\angle{APB}=\angle{BFD}=a=>APQF$ is cyclic. $\angle{QAP}=\angle{DFP}=180-2a=>\angle{AQP}=a=>AP=AQ$ EDIT. It seems some of our readers think this was not only unneeded, but spammish. Indeed, why take the trouble to correct some patently wrong mathematical statements, or mention the similarity with some already posted solution? Some of our users like it muddy, the same their own thinking processes are ...

Quite a simple angle chase. Can't believe this made an IMO shortlist!

Since : $\angle{PQF}=\angle{BQD}=\pi-\angle{PBC}-\angle{FBD}=\pi-\angle{PAC}-\angle{BAC}=\pi-\angle{PAF}$. Hence $APQF$ is cyclic, and then $\angle{AQP}=\angle{AFE}=\angle{ACB}=\angle{APB}=\angle{APQ}$, and so we are done !

Here is my solution: Let $O$ be the circumcenter of $\triangle ABC$. Let the other intersection of $EF$ with the circumcircle be $R$. Then it is well known that $AO \perp EF$, so $AP=AR$. Thus $\angle PBA= \angle ABR$. Also $\angle AFR= \angle BFD=\angle AFQ$, so $\angle BFR= \angle BFQ$. So by $ASA$ congruency, $\triangle BFQ \cong \triangle BFR$. Then $\triangle BAQ \cong \triangle BAR$. So $\triangle BAQ$ and $\triangle BAR$ have congruent circumcircles. Since $AP$ and $AQ$ are both intercepted by $\angle ABQ$ in these circumcircles, $AP=AQ$.

Looks projective... Lemma 1: $PAFQ$ is cyclic, with circle $w$. $\angle APQ = \angle APB = \angle ACB = \angle BFD$ where the last follows from anti-parallel $DF$ wrt $\angle ABC$. Lemma 2: Let $FC \cap BP = K$, then $(P, Q; K, B)$ is harmonic. Let $EF \cap BC = X$, so $(X, D; C, B) = -1 \implies F(X, D; C, B) = -1 \implies (P, Q; K, B) = -1$. Proof: Take the pencil $F(P, Q; K, B)$ and intersect it with $w$ to get $PQA'A$ is a harmonic quadrilateral, where $A' = FK \cap w$. Since $\angle KFB = 90 \implies KF$ bisects $\angle PFQ$, so $A'$ is the midpoint of arc $PQ$, and thus $A$ is the midpoint of the supplementary arc $PQ$, so $AP = AQ$. Truth be told, Jutaro's is the one I saw first

Take $P'$ the second point of intersection of $EF$ with the circle. We will prove that $Q$ is the mirror image of $P'$ about $AB$. To do this, we take $Q'$ the mirror image of $P'$ about $AB$, and try to prove that is equivalent to $Q$. Firstly, we show that $Q' \in BP \Leftrightarrow \angle ABQ' \equiv \angle ABP \Leftrightarrow \angle P'BA \equiv \angle ABP \Leftrightarrow \angle P'A \equiv \angle PA$ (as arcs of circles). But this is equivalent to $EF || t_A$ (the tangent through $A$ at the circumcircle), which is obvious ($\angle AFE \equiv \angle ACB$). Now, to prove that $Q' \in FD$: $Q' \in FD \Leftrightarrow \angle BFQ' \equiv \angle BFD \Leftrightarrow \angle BFP' \equiv \angle BFD \equiv \angle EFA$, which is obvious. So, $Q \equiv Q'$. Since $Q$ and $P'$ are symmetric, we have $AP' = AQ$. Since $\angle AP' \equiv \angle AP$, we also have $AP' = AP$. Hence, the conclusion follows.

This is just angle chasing. Let $\angle PCB = \angle PAC = \theta.$ We have that $\angle QFB = C$ and $\angle FBP = B - \theta,$ so $\angle FQB = A + \theta = \angle BAC + \angle CAP = \angle BAP,$ so $APQF$ is cyclic. It follows that $\angle AQP = C.$ From cyclic $ABCP,$ we have that $\angle APB = C,$ so $\angle APQ = \angle AQP,$ which implies $\triangle APQ$ is isoceles with $AP=AQ.$ $\blacksquare$

This problem,I should say,is a really easy geometry problem.First of all,we have $\angle{APQ}=\angle{APB}=\angle{ACB}=\angle{BFD}=\angle{BFQ} \Rightarrow APQF$ is cyclic.Finally $\angle{AQP}=\angle{AFP}=\angle{AFE}=\angle{ACB}=\angle{APQ} \Rightarrow AP=AQ %Error. "Blackbox" is a bad command.$.

very easy for IMO due to concylicity of $APCB$ we get $\angle APQ=\angle APB=C$ now $\angle AFC=\angle ADC = 90$ so that $AFDC$ is cycic quad. which gives $\angle AFD=\angle AFQ=180-C=180-\angle APQ$ so that $AFQP$ is cyclic. which gives $\angle AQP=\angle AFP = C=\angle APQ$ and hence $AP=AQ$

Dear Mathlinkers, revisiting this problem without angles…. 1. EQ being parallel to the tangent to (O) at C, and according to the Reim’s theorem, A, P, E Q are concyclic 2. A, H, E, F being concyclic and according to an usefull lemma (http://jl.ayme.pagesperso-orange.fr/Docs/Miquel.pdf p. 23-24), A, C, H, Q are concyclic. 3. (HCHQ) being the B-Carnot’s circle and equal to (O), APQ is A-isoceles. Sincerely Jean-Louis

Surprisingly easy : $A,P,B,C$ are cocyclic , implying $APQ=APB=ACB$ And $APQ=APB=ACB=BFD=180-QFA$ yields the cocyclity of $A,F,Q,P,$ , implying that $AQP=AFE=ACB$ So $APQ=AQP$ => $AP=AQ$

We know that $APBC$ is cyclic so $\angle APQ =\angle C$ $AFDC$ and $BFEC$ are also cyclic so $\angle BFD =\angle QFA = \angle AFE = \angle C$ This implies $AFPQ$ cyclic so $\angle AQP =\angle C = \angle APQ$ This implies $AP =AQ$

In $\Delta ABC$ , $\angle BEC = \angle BFC = 90^{o}$. So, $BFEC$ is cyclic quadilateral. So, $\angle BFE + \angle BCE = 180^{o}$, $\angle BCE = 180^{o} - \angle BFE = \angle AFE$. Since, $APBC$ is cyclic. So, $\angle APB = \angle ACB$. Now, $\angle AFP = \angle BCE = \angle APB = \angle APQ$. $\angle AFP = \angle APQ$. Now to prove that $AP = AQ$, we should have to prove that $\angle APQ = \angle AQP$ i.e. $\angle AFP = \angle AQP$, which reduces to prove that $AFQP$ is cyclic. Now, $AFHE$ is cyclic. So, $\angle AFP = \angle AHE = \angle BHD$. But, $BFHD$ is cyclic. So, $\angle BHD = \angle BFD = 180^{o} - \angle AFQ$. So, $\angle AFP = \angle BHD = 180^{o} - \angle AFQ$. So, $\angle AFP + \angle AFQ = 180^{o}$. So, $AFQP$ is cyclic quadilateral. So, $\angle APQ = \angle AQP$. Hence, Proved.

Let $EF$ intersect the circumcircle of $ABC$ again at $P'$. Since $AO\perp EF$, we get $AP=AP'$, so the reflection of $P'$ over $AB$ lies on $BP$. Since $\angle AFE=\angle BFD$, the reflection of $P'$ over $AB$ also lies on $DF$, so the reflection of $P'$ over $AB$ must be $Q$. Therefore, $AP=AP'=AQ$.

WLOG, assume that $\angle B>\angle C$ and $P$ is closer to $E$ than $F$. The other configuration can be proven similarly. Assume all angles are directed. C1: I claim that $AFQP$ is cyclic. This is because $$\angle PAF+\angle PQF=(\angle A+\angle PAC)+(180-(\angle QBD+\angle QDB)),$$ and by orthocenter configurations, we have that $BDHF$ is cyclic, meaning that $$\angle QDB=\angle A,$$ and since $ABCP$ is cyclic, we also have that $$\angle QBD=\angle QBC=\angle PAC,$$ which gives that $$\angle PAF+\angle PQF=(\angle A+\angle PAC)+(180-(\angle A+\angle PAC))=180,$$ meaning that $AFQP$ is cyclic, as desired. C2: I now claim that $\angle APQ=\angle C$. This is because since $AFQP$ is cyclic by (C1), $$\angle APQ=180-\angle AFQ=\angle BFD,$$ and since $BFHD$ is cyclic by orthocenter properties, this gives us $$\angle BFD=\angle BHD=90-\angle HBD=\angle C.$$ This means that $\angle APQ=\angle C$, as desired. C3: Finally, I claim that $\angle AQP=\angle C$. Since $AFQP$ is cyclic by (C1), this gives that $$\angle AQP=\angle AFP=\angle AFE,$$ and by orthocenter properties, we have that $AFHE$ is cyclic, meaning that $$\angle AFE=\angle AHE=90-\angle HAE=\angle C,$$ meaning that $\angle AQP=\angle C$, as desired. Finally, combining (C2) and (C3), we have that $\angle APQ=\angle AQP=\angle C$, meaning that $\triangle APQ$ is isosceles with $AP=AQ$, finishing the problem.

config issues are stupid so henceforth we assume there are none because they can be dealt with easily :clown: Assume that $AB > AC$ and $P$ is on minor arc $AC$. Then, $\angle AFQ = 90 - \angle BFD = 90 - \angle BHD = \angle DBH = 90 - \angle ACB$ because $BDHF$ is clearly cyclic. Additionally, $\angle APQ = \angle APC$ because $A$, $B$, $C$, $P$ concyclic. Therefore, $AFQP$ is cyclic because $\angle APQ + \angle AFQ = 180$. To finish this problem, we note that $\angle AQP = \angle AFP = \angle AFE = \angle AHE = 90 - \angle EAH = \angle ACB = \angle APB$ so $APB$ is isosceles and we are done.

Solve after long time

Let $H$ denote the orthocentre of $\Delta ABC$. Let $\angle ACB=\theta\implies\angle APB=\theta$, as points $A,B,C,P$ all lie on $(ABC)$. Also, as $AEHF$ is cyclic, $\angle CAD=\angle EFH=90^{\circ}-\theta\implies\angle AFE=\theta$. Again, as $H$ is the incentre of $\Delta EFD$, we have $\angle BFD=\angle AFE=\theta$. Therefore, $AFDP$ is cyclic. Then $\angle AFP=\angle AQP=\theta\implies AQ=AP$, and we're done!

@above, Ig you need to use directed angles or prove for all possible configurations...

Amir Hossein wrote: Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$ Proposed by Christopher Bradley, United Kingdom $\angle PQF=\angle BQD=180-\angle QBD-\angle BDQ=180-\angle B-\angle A-\angle PBA=\angle C-\angle PBA=\angle PAF$ So $P,Q,F,A$ are cyclic. $\angle QPA=\angle QFA=180-\angle AFD=180-(180-\angle C)=\angle C=\angle AFE=\angle PQA$ Which gives $AQ=AP$

We claim that $(AFPQ)$ is cyclic, this is trivial by observing $\angle QPA = 180 - \angle BPA = \angle ACB = \angle BFD = \angle AFQ$. Now we see that $\angle AQP = 180 - \angle AFP = 180 - \angle BFD = \angle ACB = 180 - \angle BPA = \angle QPA$, so we are done.

I have discussed this problem in my EGMO YouTube tutorial ch1 angle chasing practice part

I will show the case where point $Q$ is inside of the triangle $ABC$, other situation can be shown similarly. Since $\angle APQ=\angle C$ and $\angle AFQ=90^{\circ}+\angle CFD=90^{\circ}+\angle CAD=180^{\circ}-\angle C$, we have that the points $A$, $P$, $F$ and $Q$ are concyclic. Now, observe that we wish to show $\angle APQ=\angle AFP$ which is equilavent to $\angle C=\angle AFP=90^{\circ}-\angle CAD=\angle C$ as desired.

Since $APBC$, $BFEC$ and $AFDC$ are cyclic, we have that $\angle AFQ = \angle BFD = \angle ACD = \angle ACB = \angle APQ$, which means that $AFPQ$ is cyclic. Then, because of the above cyclic quadrilaterals, we have that $\angle AQP = \angle AFE = \angle ACB = \angle AFQ = \angle APQ$, giving that triangle $APQ$ is isosceles. The above proof is for $Q$ outside triangle $ABC$. An analogous proof follows for $Q$ inside $ABC$.

Assume $P$ is closer to $E$, the other config is the same solution. $AFQP$ is cyclis because $\angle QPA = 180 - \angle QFA$. Now $\angle AQP = \angle AFP = \angle AFE = \angle C$ but $\angle APQ = \angle C$ as well.

Suppose that $APQF$ forms a non-self-intersecting quadrilateral. Then $\angle APQ = \angle ACB = \angle AFE$ so it suffices to show that $APQF$ is cyclic. But this is easy since $\angle APQ = \angle ACB = 180^\circ - \angle AFD,$ completing the proof. Note that the case when $AQPF$ is cyclic can be dealt with similarly. QED