Solution by TomTom314: Let $F'$, $B'$, the midpoints of $BP$, $CP$ and $K'$ the circuncenter of $BCP$. Then we can show, $|EE'|=|FF'|=\frac{1}{2}|AP|$ and both lines are perpendicular to $BC$ using the Intercept theorem. Because $EK$ , $FK$ are parallel to $E'K'$, $F'K'$ we conclude that $KK'$ is perpendicular to $BC$ and therefore $K$ lies on the perpendicular bisector of $BC$. [asy][asy] import olympiad; size(7 cm); defaultpen(fontsize(10pt)); pair A, B, C, H, E, F, P, M, G, K, O1, O2, O3; A = (2,5); B = (6,0); C = (0,0); E= (1, 2.5); F= (4,2.5); P= (2,3); O1 = (P+C)/2; O2 = (P+B)/2; M = orthocenter(C,E,P); H=orthocenter(A,B,C); G= orthocenter(P,B,F); K= extension(E,M,F,G); O3 = (A+P)/2; draw(circumcircle(A,B,C), dashed+blue); draw(A--B--C--A); draw(B--K); draw(C--K); draw(E--O1, lightblue); draw(F--O2, lightblue); draw(A--O3, lightblue); dot("$A$",A);dot("$B$",B);dot("$C$",C);dot("$F$" ,F, blue);dot("$E$" ,E, blue);dot("$P$",P, purple);dot("$K$", K, red);dot("$B'$", O1, orange); dot("$F'$", O2, orange);dot(O3); [/asy][/asy]