Let $ABC$ be an acute and scalene of circumcircle $\Gamma$ and orthocenter $H$. Let $A_1,B_1,C_1$ be the second intersection points of the lines $AH, BH, CH$ with $\Gamma$, respectively. The lines that pass through $A_1,B_1,C_1$ and are parallel to $BC,CA, AB$ intersect again to $\Gamma$ at $A_2,B_2,C_2$, respectively. Let $M$ be the intersection point of $AC_2$ and $BC_1, N$ the intersection point of $BA_2$ and $CA_1$, and $P$ the intersection point of $CB_2$ and $AB_1$. Prove that $\angle MNB = \angle AMP$ .
First of all, it's easy to see by angle-chasing that $A_2, B_2, C_2$ are just the points diametrically opposite to $A, B, C$ on $\Gamma.$ Let $\omega_1, \omega_2, \omega_3$ denote the circles centered at $M, N, P$ with radii $MA, NB, PC$, respectively. Then, $\omega_1$ is tangent to $AC$ and goes through $A$ and $B$, with cyclically similar results for the other two circles. Let $X$ be the intersection of $\omega_1$ and $\omega_2$ which isn't $B$. We then have from the tangent condition that $\angle XAC = \angle XBA = \angle XCB.$ But then $\angle XAC = \angle XCB$ implies that $X$ is on $\omega_3$ as well, hence $\omega_1, \omega_2, \omega_3$ all share the common point $X$. Now, observe that since $MB = MX, NB = NX$, we have that $MXNB$ is a kite, hence $\angle MNB = \angle XNM = \frac12 \angle XNB = \angle XCB.$ Analogously, we have that $AMXP$ is a kite, hence $\angle AMP = \angle XMP = \frac12 \angle AMX = \angle ABX.$ Hence, as $\angle ABX = \angle BCX$ is known, we're done.
$\square$
Remark. $X$ is known as the $n$th Brocard point, where $n \in \{1, 2\},$ I can't remember which.