Where Do They Concur?

Let $A'$ be the midpoint of segment $BC$ . By Brokard's Theorem on cyclic quadrilateral $BCEF$ , $A'$ is the orthocenter of triangle $AHP$ . Therefore, $l_A$ passes through $A'$ , i.e. $l_A$ passes through the centroid of triangle $ABC$ . Similarly, $l_B$ and $l_C$ pass through the centroid of triangle $ABC$ , so $X$ is the centroid of triangle $ABC$ . $\blacksquare$

This is my first solution on Brilliant, so please give me feedback.

Image link: http://s30.postimg.org/wy0qtjcrl/Untitled.png

Let $M$ be the midpoint of $BC.$ We shall show that $AM= \ell_A.$ Let $J$ be the foot of perpendicular from $H$ on $AM.$ It shall suffice to show that $P,H,J$ are collinear.

Since $HF \perp AB, HJ \perp AJ,$ and $HE \perp AC,$ points $A,F,H,J,E$ lie on a circle. Let $\omega _1$ denote this circle. Also, since $BE \perp AC$ and $BF \perp AB,$ points $B,C,E,F$ lie on a circle. Let $\omega_2$ denote this circle. Finally, let $\omega_3$ denote the nine point circle of $\triangle ABC,$ which passes through the points $M,D,E,F.$

Observe that $EF,BC,$ and $HJ$ are the radical axes of $(\omega_1, \omega_2), (\omega_2, \omega_3),$ and $(\omega_3, \omega_1 )$ respectively, so they must concur at the radical center of $\omega_1, \omega_2, \omega_3.$ It follows that $P,H,J$ are collinear. $\blacksquare$

Analogous arguments show that $\ell_B$ and $\ell_C$ are the medians passing through $B,C$ respectively. Hence, $\ell_A, \ell_B, \ell_C$ are concurrent at the $\fbox{centroid}$ of $\triangle ABC.$