, 2021

Solution 1. Let $X$ be a point on $BC$ such that $LX \perp AB$, as seen in figure 18. It is enough to prove that
$$\frac{DP}{PL} = \frac{DK}{KX}$$
because then $PK \parallel LX$ and $LX \perp AB$.
Applying Menelaos for triangle BDL and transversal MPC we get
$$\frac{DP}{PL} \cdot \frac{LM}{MB} \cdot \frac{BC}{CD} = 1,$$
and Menelaus for triangle BLC and transversal AMD gives
$$\frac{BM}{ML} \cdot \frac{LA}{AC} \cdot \frac{CD}{DB} = 1.$$
Multiplying these two equalities yields
$$\frac{DP \cdot BC \cdot AL}{PL \cdot BD \cdot AC} = 1.$$
Note, however, that $AL = AD = AC \sin \gamma$, $BD = AB \cos \beta$, and, by the sine rule, $\frac{AB}{BC} = \frac{\sin \gamma}{\sin \alpha}$, where $\alpha = \angle BAC$, $\beta = \angle CBA$ and $\gamma = \angle ACB$. Therefore
$$\frac{DP}{PL} = \frac{BD \cdot AC}{BC \cdot AL} = \frac{AB \cos \beta \cdot AC}{BC \cdot AC \sin \gamma} = \frac{\sin \gamma \cos \beta}{\sin \alpha \sin \gamma} = \frac{\cos \beta}{\sin \alpha}.$$
On the other hand, since $DK = KL$, $\angle KLX = \pi - \alpha$, and $\angle LXX = \frac{\pi}{2} - \beta$, we have by the sine rule

Therefore
$$\frac{DP}{PL} = \frac{\cos \beta}{\sin \alpha} = \frac{DK}{KX}$$

Solution 2. Let $\omega$ be the circle with center $K$ an radius $KD$, as in figure 19. Then $\omega$ is tangent to $AD$ and $AL$. Let $BC$ intersect $\omega$ at $D$ and $Q$. Let $BM$ intersect $\omega$ at $L$ and $R$. Let $QP$ intersect $BL$ at $S$.
Cross-ratio chasing gives, through the projections $BL \to D$-pencil $\to \omega \to L$-pencil $\to BC \to P$-pencil $\to BL$,
$$\begin{aligned} (L, R; M, B) &= (DL, DR; DM, DB) = (L, R; D, Q) = (LC, LB; LD, LQ) \\ &= (C, B; D, Q) = (PC, PB; PD, PQ) = (M, B; L, S) = (L, S; M, B), \end{aligned}$$
therefore $R = S$.
It is clear now that $P$ lies on the polar lines of both $A$ and $B$ with respect to $\omega$, therefore $AB$ is the polar line of $P$. This implies that $PK \perp AB$.