Note that since the orthocenter $H$ is the incenter of $\triangle{DEF}$, $EB$ bisects $\angle{DEP} \implies$ if $ED = EP$, $BDEP$ is a kite, and so is $KDBP$ if $K = EB \cap (ABC)$. This also means $BK$ bisects $\angle{PBC}$, so $KP = KC = KD \implies$ the line through $K$ and the midpoint of $CD$ is perpendicular to $BC$. Thus, if $M$ is the midpoint of $AC$, $KM \parallel AD \implies KMHA$ is a rhombus, which means $AE = EM$, or $AE : EC = 1 : 3$. Note that $PD \parallel AC$, so $CQ \parallel DP$ if and only if $Q \in AC$. $BP \cap AC$ is the reflection $C'$ of $C$ over $E$, and $DF \cap AC$ is the point on ray $CA$ which forms a harmonic bundle with $A, E, C$ by Ceva-Menelaus, so it suffices to show that $(C', E; A, C) = -1$ which is obvious since $\frac{AE}{AC'} = \frac{CE}{CC'} = \frac{1}{2}$. $\square$

First, let $H$ be the orthocenter and $\angle CAD=\angle EBC=\alpha$, $ \angle DAB=\angle FCB=\theta $ and $\angle FCA=\angle EBA=\beta$, and $\alpha + \beta + \theta=90^\circ$. Let $T = EF \cap (ABC)$, different from $P$. Since $HEAF$ is a cyclic quadrilateral $\angle BEP =\angle HEF = \angle HAF = \angle DAB = \theta$, same way, $HDCE$ is cyclic and $\angle FCB= \angle HCD= \angle HED= \angle BED=\theta$, so $EB$ bisects $\angle DEP$, analogy $AD$ bisects $\angle FDE$ with $\angle FDE=2\beta$. Let $EB$ and $DP$ intersect at $K$. If $ED=EP$ then $EKP=90^\circ $, then $AC \parallel DP$. Since $AFDC$ is cyclic then $\angle BFD= \angle DCA = \beta + \theta$. Also since $EFBC$ is cyclic $\angle PFB= \angle AFE = \angle ECB = \beta + \theta$. Since $DP \parallel AC$ we have that $\angle ACB= \angle PDB = \beta + \theta$ therefore, since $\angle PFB= \angle PDB$, $PFDB$ is cyclic. Since $KDE=90^\circ$ we have that $\angle PDF=\alpha - \beta$, son $\angle ABP = \alpha - \beta$. Arc $\frac{AB}{2}=\alpha+\theta=\angle PFB=\frac{PB}{2}+\frac{AT}{2}$, from here $\frac{AT}{2}=\frac{AP}{2}=\alpha - \beta$. Also $\angle PAB + \angle PBA= \beta+ \theta=\angle PBF + \angle BQF$ since $\angle QFA= \angle BFD$. Therefore $\angle PQF = \angle PAF$ and $PQAF$ is cyclic. Since $\angle APF = \alpha - \beta = \angle AQF = \angle PDF$ we have that $AQ \parallel DP$. From here it follows that since $AC \parallel AQ \parallel DP$, so $A, Q, C$ are collinear and $CQ \parallel DP$

With angle chasing, one can find that $AC\parallel DP,$ and $B, P, F, D$ are concyclic points. Then, by Radical Axis Concurrence Theorem on the circles $(ABC),$ $(BPFD),$ and $(AFDC)$, we get that the lines $BP, DF,$ and $AC$ are concurrent, which means that $Q$ lies on $AC$. Combining with $AC\parallel DP,$ the problem is finished.

My sol. is similar to #4. From simple angle chasing we have $\angle FBD=\angle FPD$, which means that $(P F H B D)$ is cyclic. Then from Radical axes theorem in circles $(A F D C)$, $(P F H D B)$ and $(A P B C)$, we have $PB$, $FD$ and $AC$ concurrent which means that $Q$, $A$, $C$ are collinear. From $(B P F H D)$ cyclic we have $\angle HPB=90$ and we know that $HB$ perp. to $PD$ (because of $EB$ is bisector of $\angle PED$), then $\angle BPD=\angle PHB$ $(1)$. Then $90=\angle BPH=\angle QPH$ and $\angle QEH=90$, which means that $(Q P H E)$ is cyclic, then $\angle PHB=\angle BQC$ $(2)$. From $(1)$ and $(2)$ we have $\angle BPD=\angle BQC$, which completes the solution (I haven't seen that $AC//DP$:(, so my solution is longer than #4).

By angle chasing \[\angle PED = 180^{\circ} - 2 \angle B \Rightarrow \angle EPD = \angle B.\]Notice that $\angle FPD = \angle EPD = \angle B \Rightarrow FPBDH$ is cyclic. Since $PBFD$, $FDAC$ and $PBAC$ are cyclic, by Radical Axis Theorem $PB$, $FD$ and $AC$ are concur at $Q$. Finally \[\angle QEP = \angle AEF = \angle B = \angle EPD\]thereby $CQ \parallel DP$ $\blacksquare$

$EP = ED$ and $EB$ is the angle bisector of $\angle PED$, so, $\triangle EPB \equiv \triangle EDB \implies \angle EPB = \angle BDE = \angle FDC \implies FPBD$ is cyclic. Therefore, $P$ is the miquel point of $ACDF$, which is cyclic, so, $BP$, $DF$, $AC$ meet at $Q$ (just power center). Now, it is easy to see that $\triangle BEC \equiv \triangle BEQ$ ($ASA$), then, $BC = BQ$, but we already knew that $BD = BP$ (first congruence) $\implies DP \parallel CQ.$

AlanLG wrote: Let $ABC$ an acute triangle and $D,E$ and $F$ be the feet of altitudes from $A,B$ and $C$, respectively. The line $EF$ and the circumcircle of $ABC$ intersect at $P$, such that $F$ it´s between $E$ and $P$. Lines $BP$ and $DF$ intersect at $Q$. Prove that if $ED=EP$, then $CQ$ and $DP$ are parallel. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $H$ be the orthocenter of $\triangle ABC$ Since $EB$ is the angle bisector of $\angle DEP$ and $\triangle DEP$ is isosceles: $$\Rightarrow EH\perp PD$$$$\Rightarrow BE\perp PD...(\alpha)$$$$\Rightarrow PD//AC...(I)$$BIGFLIPPA noted that 2010 ISL G1 makes the problem trivial By 2010 ISL G1: $$\Rightarrow AQ=AP$$$$\Rightarrow \angle AQP=\angle APQ$$Since $PACB$ is cyclic: $$\Rightarrow \angle AQP=\angle APQ=\angle ACB$$Since $CDHE$ is cyclic: $$\Rightarrow \angle AQP=\angle ACB=\angle AHE=\angle BHD...(\beta)$$By $(\alpha)$ and $(\beta)$: $$\Rightarrow \angle AQP=\angle BDP...(\theta)$$Since $BFEC$ is cyclic: $$\Rightarrow \angle AQP=\angle APQ=\angle ACB=\angle AFE$$$$\Rightarrow \angle AQP=\angle AFE$$$$\Rightarrow AFPQ \text{ is cyclic}$$$$\Rightarrow \angle AFE=\angle APQ=\angle AQP=\angle AFQ$$$$\Rightarrow \angle AQP=\angle PFB=\angle BFD...(\omega)$$By $(\theta)$ and $(\omega)$: $$\Rightarrow BPFD \text{ is cyclic}$$$$\Rightarrow \angle BPD=\angle AQP=\angle BFD=\angle BPD...(\epsilon)$$By $(I)$ and $(\epsilon)$: $$\Rightarrow Q, A \text{ and }C \text{ are collinears}$$$$\Rightarrow CQ//DP_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

It is known that the altitude $BE$ is a bisector of $\angle FED$. Therefore, if $ED=EP$, then $\triangle PED$ is isosceles and, more importantly, $PD\perp BE$ because $BE$ is a perpendicular bisector of DP. We can also obtain that $BPD$ is isosceles with $BP=BD$ $PD\perp BE$ implies that $PD\parallel AC$. So $CQ\parallel DP \iff Q\in AC$. That means that we must prove: $BP\cap DF\cap AC=Q$. $ACDF$ is cyclical, $\Rightarrow \angle ACB=\angle BFD$. $PD \parallel AC \Rightarrow \angle ACB=\angle PDB$, and since $BPD$ is isosceles, $\angle ACB=\angle PDB=\angle BPD$. So $\angle BFD=\angle BPD \Rightarrow PFDB$ is cyclical. Finally, $BP, DF, AC$ are radical axis of the pairs of circles $PFDB, ACDF,ACBP$, proving that $BP, DF, AC$ concur. And since $BP\cap DF=Q$, then $BP\cap DF\cap AC=Q$.

We show a way to prove that $PD//AC$ with projective geometry since it looks like the other solutions prefer to prove it by angle chasing, but it's hard to make the conjecture that $Q$ lies on $AC$ without a good diagram, and here we show the motivation why we would want to prove that $Q$ lies on $AC$ which can then be proven by first proving by angle chasing that $[BDHFP]$ is cyclic with $H$ being the orthocentre of triangle $ABC$ and then finish the problem with radical centre like some other solutions. Let $S = EB \cap PD$, by simple angle-chasing we know that $\angle PEB=\angle DEB$, so $S$ is the midpoint of $PD$. Let $T = BC \cap EF$ By Ceva-Menelaus, we know that $(T, D; B, C) = E(T, D, B, C) = E(P, D; S, PD \cap AC) = (P, D; S, PD \cap AC) = -1$, since $S$ is the midpoint of $PD$, $PD \cap AC$ is a point at infinity, thus we have $PD//AC$, as desired.

hectorleo123 wrote: BIGFLIPPA noted that 2010 ISL G1 makes the problem trivial By 2010 ISL G1: yea

I'm the author of this problem, Miguel Ángel Hernández from the Dominican Republic. I hope you enjoyed solving my problem. I came up with this idea while I was solving ISL 2010 G1 from Evan Chen´s textbook, for training purposes. I noticed that QPDC is cyclic and then thought if I added some extra point or condition I could exploit the potential of that cyclic to discover some nice property. After some experimentation, I arrived to the condition EP=ED. Here is my solution.

Proof: Let $H$ be the orthocenter. Then $\angle FEB=\angle FAH=\angle FCB=\angle DEB.$ And $EP=ED, EB=EB.$ Therefore, $\triangle EPB \cong \triangle EDB.$ So $\angle EPB=\angle EDB=\pi-\angle EDC=\pi-\angle EHC=\pi-\angle FHB=\pi-\angle FDB=\angle FDC$. Therefore, $P,B,F,D$ are concyclic. Obviously, $A,F,D,C$ are concyclic. According to the properties of the radical center, the radical axis of $\odot(ABC)$ and $\odot(AFD)$ is $AC$, the radical axis of $\odot(ABC)$ and $\odot(PBD)$ is $PB$, the radial axis of $\odot(PBD)$ and $\odot(FDC)$ is $DF$, and these three lines intersect at point $Q$. So $A,Q,C$ are collinear. At this point, from $\angle PBE=\angle DBE$ and $BE \perp AC$, we know that $BQ=BC$. And from $\triangle EPB \cong \triangle EDB$, we have $BD=BP$. Therefore, $CQ$ and $DP$ are parallel. Q.E.D.

Firstly, we must notice that $PD//AC$. Call $\angle ABC= B$. If $H$ is the othocenter, then $AEFH$ is cyclic, because $\angle AEH+\angle AFH= 90+90= 180$. So, $\angle FEH= \angle HAF= \angle DAB= 90-B.$ Similarly, $ECDH$ is cyclic and $\angle HED= 90-B$. So, $EB$ is angle bisector of $\angle FED= \angle PED$. Also, as $EP= ED\implies PD\perp EB\perp AC$. So, $PD//AC$. Notice that $FDCA$ is cyclic, because $\angle AFC= \angle ADC= 90$ Also, from $EP= ED\implies \angle EDP=\angle EPD= B= \angle FBD$. So, $FPBD$ is cyclic. Then, from radius center in $(ABC), (PFBD), (FDCA), BP, DF$ and $AC$ concur. As $Q= BP\cap DF$ and $BP, DF, AC$ concur $\implies Q\in AC\implies CQ\in AC\implies DP//CQ$, as desired.

First, as $BE$ bisects $\angle PED$, we have $PEDB$ is a kite, so $BP=BD$. Then we got $\measuredangle FPB=\measuredangle EPB=\measuredangle BDE=\measuredangle CDE=\measuredangle FDB$, so $FDPB$ is cyclic. Radical center on $(FDBP), (FDCA), (ABC)$ gives $BP,DF,AC$ concurrent, so $Q\in AC$. Now angle chasing gives $\measuredangle EPB=\measuredangle FPD=\measuredangle FBD=\measuredangle ABD=\measuredangle AE=\measuredangle CED$, so $DP\parallel AC$.