parmenides51 wrote: A point $E$ lies on the extension of the side $AD$ of the rectangle $ABCD$ over $D$. The ray $EC$ meets the circumcircle $\omega$ of $ABE$ at the point $F\ne E$. The rays $DC$ and $AF$ meet at $P$. $H$ is the foot of the perpendicular drawn from $C$ to the line $\ell$ going through $E$ and parallel to $AF$. Prove that the line $PH$ is tangent to $\omega$. (A. Kuznetsov) Construct rectangle $EABN$. Claim. $\overline{HN}$ is tangent to $\odot(ABE)$. (Proof) Observe that $CEHN$ is cyclic, so $\measuredangle HNE=\measuredangle HCE=\measuredangle AFB=\measuredangle EBN$ as desired. $\blacksquare$ Move point $C$ along line $\overline{NB}$. Fix line $\ell'$ tangent to $\omega$ at $N$. Point $H$ lies on $\ell'$ for all $C$. Let $P_1\overset{\text{def}}{=} \overline{AF} \cap \ell'$ and $P_2\overset{\text{def}}{=} \overline{CD} \cap \ell'$; then we want $P_1 \equiv P_2$. Note that $C \mapsto F \mapsto P_1$ is a projectivity, while $C \mapsto P_2$ is an affine mapping. Consequently, it suffices to verify $P_1 \equiv P_2$ for three positions of $C$. Meanwhile, $C \in \{B,N\}$ are clearly true. Now let $C$ be the mid-point of $\overline{BN}$. Let $O$ be the center of $\omega$ and $L=\overline{AF} \cap \overline{BN}$, then $$(\overline{AF}, \overline{PL}) \overset{C}{=} (A,E;D,\infty)=-1$$so $\overline{BN} \perp \overline{OP}$ alongside $L \in \overline{BN}$ ensures that $\overline{BN}$ is the polar of $P$ in $\omega$. Hence the claim is proved. $\blacksquare$

A bit lower-powered. Proceed as above to get $HN$ tangent to $\omega$. So we need to show $PN$ tangent $\omega$. But note: $\angle CPF = \angle FAB = \angle CNF$, so $CNPF$ is cyclic. Then $\angle BNP = \angle CFA = \angle EBA = \angle BEN$. So the tangency follows. $\blacksquare$

Let $BC \cap \odot (ABE)=K \implies ECKH$ is cyclic. Since, $\angle AFK$ $=$ $90^{\circ}$ $\implies$ $KCFP$ is cyclic $\implies $ $\angle FKP$ $=$ $\angle FCP$ $=$ $\angle DCE$ $=$ $\angle FEK$ $\implies $ $PK$ tangent to $\odot (ABE)$. But, $\angle CKP$ $=$ $\angle LFC$ $=$ $90^{\circ}-\angle ECH$ $=$ $180^{\circ}-\angle CKH$ $\implies$ $K \in PH$

Let BC meet circle at S. we will prove PH is tangent to circle at S. ∠CSE = 90 = ∠CHE ---> CSHE is cyclic. ∠PSC = ∠HEF = ∠EFA = ∠EBA = ∠SEB ---> PS is tangent to circle.