$M$ is the midpoint of line segment $BC$. $N$ and $P$ are points in $\triangle ABC$ such that $PM \parallel AB$ and $MN \parallel CA$, $CP \perp PM$ and $MN \perp NB$. $K$ is the midpoint of $NP$. $L$ is the reflection of $K$ in the circumcircle of $\triangle MNP$. Prove that $LB = LC$.
This comes from a test by my teacher where I couldn't solve any of the problems. I suppose this problem can be solved by using perspective geometry, which I only have studied the basics.
Let $\omega$ denote $(MNP)$, and let its circumcentre be $O$.
Note that if $H$ is the orthocentre of $ABC$, then $BN$ and $CP$ concur at $H$ as $MN \parallel AC$, and $MP \parallel AB$ so $BN \perp AC$ and $CP \perp AB$.
Therefore, since $\angle HNM = \angle HPM = 90^{\circ}$, $H$ is the antipodal point of $M$ in $\omega$.
Now, since $L$ is the reflection of $K$ in $\omega$, then $LN$ and $LP$ are tangent to $\omega$. This follows from similar triangles: $OK \times OL = ON^2 \implies \frac{OK}{ON} = \frac{ON}{OL}$, and $\angle LON$ is shared, so $\triangle OKN \sim \triangle ONL \implies \angle ONL = \angle OKN = 90^{\circ}$ as $K$ is the midpoint of $NP$. Therefore, $LN$ is tangent to $\omega$. A similar argument holds for $LP$.
Suppose $P_{\infty}$ is the point at infinity on line $BC$. Then, we have
$$(B, C; M, P_{\infty}) = -1.$$
Projecting these through $H$ onto $\omega$ yields
$$(N, P; M, H') = -1,$$
where $H'$ is the point on $\omega$ such that $HH' \parallel BC$. From this relation, we get $MNH'P$ is a harmonic quadrilateral, which implies $MH'$ is the symmedian of $\triangle MNP$. However, $ML$ is the symmedian of $MNP$ as $L$ is the intersection of the tangents at $N$ and $P$, so we must have $L, H', M$ collinear.
Therefore, since $H'M \perp BC$ (as $\angle HH'M = 90^{\circ}$ and $HH' \parallel BC$), then $LM \perp BC$, so $LB = LC$ as $LM$ is a perpendicular bisector of $BC$.
Extend BN to cut AC at B’. Do the same to get C’. Then, BB’ and CC’ are altitudes of $\triangle ABC$ and they will meet at H, the orthocenter. Hence A(H)A’ is the third altitude. Note that M, N. H, P lie on the same circle whose center is O and with HM as the diameter.
Let LM cut circle MNHP at S. Let SP produced cut SC at T.
Produce PO to cut circle MNHP at Q such that P(O)Q is another diameter of that circle. Note that, after such construction, HQMP is a rectangle circumscribed by the circle MNHP.
The inversion point L assures that (1) L, K, O are collinear; and (2) LN and LP are tangents to the circle MNHP.
Because of many right angles meet at P, we can ontain many equal angles. Those coded with the same color are actually equal. By simple subtraction, we get the pink coded angles are actually equal to those coded in red. Hence, $\angle A’PT = 90^0$.
$\angle HST = 90^0 + \angle blue = \angle A’PT + \angle PA’T = \angle PTC$. This means (1) HS // BC and further (2) $\angle SMC = \angle HSM = 90^0$. Required result follows.
