Triangles $ODE$ and $ODF$ are equilateral and now quick angle chase shows $BD \perp CP$ and $CD \perp BP$, so $D$ is the orthocenter of $BCP$, whence $PD \perp BC$.

By construction, $DO = DF = DE$. since $OD$ and $FE$ bisects each other, $OEDF$ is a parallelogram $\Rightarrow \angle DOF = 90 - \angle OFE = 90 - \angle FED = 90 - \frac{1}{2} \angle DOF \Rightarrow \angle DOF = 60 \Rightarrow \angle EOF = 120 = \angle BOC \Rightarrow \angle EBF = 60 = \angle PEC$, and $PE = PC \Rightarrow \triangle PEC$ is equilateral $\Rightarrow \angle OPF = 30 = OEF$, so the quadrilateral $PEOF$ is cyclic. Hence $DE = DF = DP$ so $D$ is the circumcenter of $\triangle PEF$. Let $L$ be the foot of $P$ from $EF$. We know that $\angle EPL = \angle KPC$, where $K$ is $PD \cap BC$. Is suffice to prove that $\angle PEF = \angle PCK$. Note that $\angle PEF = \angle PEC + \angle CEF = 60 + \angle ECB = \angle PCK$, so we are done. $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \blacksquare$ [asy][asy] import graph; import math; import olympiad; import geometry; import cse5; size(8cm); void my_dot(pair A) {fill(circle(A, .01),black); } pair A = dir(90); pair B = dir(210); pair C = dir(330); pair O = circumcenter(A,B,C); pair D = dir(290); real r1 = circumradius(A,B,C); path c1 = circle(O, r1); path c2 = circle(D, r1); pair E = intersectionpoints(c1,c2)[1]; pair F = intersectionpoints(c1,c2)[0]; pair P = extension(B,F,C,E); pair L = foot(P,F,E); pair K = extension(P,D,B,C); draw(circumcircle(A,B,C)); draw(A--B--C--cycle, linewidth(1.2)); draw(B--P); draw(E--P); draw(O--F); draw(O--E); draw(F--E); draw(P--L); draw(P--K); my_dot(A); my_dot(B); my_dot(C); my_dot(O); my_dot(D); my_dot(E); my_dot(F); my_dot(P); my_dot(L); my_dot(K); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, dir(340)); label("$O$", O, NE); label("$D$", D, SE); label("$E$", F, dir(SW)); label("$F$", E, dir(E)); label("$P$", P, dir(340)); label("$L$", L, dir(245)); label("$K$", K, SE); [/asy][/asy]

Notice that $OEDF$ is a rhombus therefore $\angle EDF=120^\circ$ and $\angle EAF=60^\circ$. Hence $BE=CF$. Also notice that $$\angle PEF= 60^\circ+\angle CAF=60^\circ+\angle BCE=\angle AFE$$ So $AE\parallel PF$. Similarly $AF\parallel PE$. Therefore $AEPF$ is a parallelogram. Notice that $\angle EBD+\angle BEA=30^\circ+60^\circ=90^\circ\implies BD\perp PC$. Similarly $CD\perp PB$. Hence $D$ is the orthocenter of $\triangle PBC$.

This problem was proposed by me, Santiago Rodriguez from Colombia. I wrote it as a challenge, as I wanted to see if I could create a non-trivial problem about an equilateral triangle. This is what I came up with, I believe it is somewhat cute. My solutions are quite similar to the ones already posted so I won't post them unless I'm asked to. I hope that you enjoy my problem.

I have attached a diagram for supporting my argument!

lifeismathematics wrote:

I have attached a diagram for supporting my argument! beautiful!

Es un Resultado conocido que las intersecciones de la mediatriz de un rádio con la circunferencia forman triangulos equilateros con los puntos que determinan el radio => ∆ABC~∆OFD~ODE Declaración: D es el ortocentro del triángulo BCP Demostración: Vamos por números complejos: La condición ∆ABC~OFD es equivalente a: a--b/c--b=f/d(¶) => si BD es perpendicular a CF => debe cumplirse: c--f/d-b + c--f(db)/d--b(cf)=0 <=> c--f/d--b • [(db/cf) +1]=0 Pero por (¶) sabemos que d/f=c--b/a--b Sustitúyendo llegamos a la conclusión de que si probamos que (c--b)b/(a--b)c=-1 concluimos abriendo y cancelando terminos semejantes obtenemos que es equivalente a probar que ac=b² lo cual es cierto ya q el ∆ ABC es equilátero => concluimos, de forma análoga probamos que CD es perpendicular a BP y en consecuencia que D es el ortocentro de ∆BCP => PD es perpendicular a BC.