Ukrainian National Mathematical Olympiad

Denote $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$. Let $N$ be the midpoint of arc $ACB$ of the circumcircle of $\triangle ABC$ (Fig. 35). We are going to show that $N$ belongs to the line $DE$.

Firstly, observe that both points $D$ and $E$ lie inside $\triangle ANB$. Indeed, since both of them are inside of $\triangle ABC$, $\frac{1}{2}\beta = \angle EAC < \angle BAC = \alpha$ and $\frac{1}{2}\alpha = \angle DBC < \angle ABC = \beta$.

Also $\angle NAB = \angle NBA = 90^\circ - \frac{1}{2}\angle ACB = 90^\circ - \frac{1}{2}\gamma = \frac{1}{2}(\alpha + \beta)$,
$$\angle EBA = \frac{1}{2}\beta < \frac{1}{2}(\alpha + \beta) = \angle NBA \text{ and } \angle EAB = \alpha - \frac{1}{2}\beta < \frac{1}{2}(\alpha + \beta) = \angle NAB,$$
because $\frac{1}{2}\alpha < \beta$. So $E$ lies inside $\triangle ANB$, similarly $D$ lies there. Now we have that
$$\begin{aligned} \angle EAB &= \alpha - \frac{1}{2}\beta, \\ \angle NAE &= \frac{1}{2}(\alpha + \beta) - (\alpha - \frac{1}{2}\beta) = \beta - \frac{1}{2}\alpha, \\ \angle NBI &= \frac{1}{2}\alpha, \quad \angle IBA = \frac{1}{2}\beta. \text{ By Ceva's Theorem from the triangle } ABN \text{ and the point } E \end{aligned}$$
Theorem from the triangle $ABN$ and the point $E$
$$1 = \frac{\sin \angle ANE}{\sin \angle ENB} \cdot \frac{\sin \angle NBE}{\sin \angle EBA} \cdot \frac{\sin \angle BAE}{\sin \angle EAN}$$
After substitutions we get
$$\frac{\sin \angle ENB}{\sin \angle ANE} = \frac{\sin \frac{1}{2}\alpha}{\sin \frac{1}{2}\beta} \cdot \frac{\sin(\alpha - \frac{1}{2}\beta)}{\sin(\beta - \frac{1}{2}\alpha)}$$
Analogously, $\angle DAB = \frac{1}{2}\alpha$,
$$\begin{aligned} \angle DAN &= \frac{1}{2}\beta, & \angle DBA = \beta - \frac{1}{2}\alpha, \\ \angle NBD &= \alpha - \frac{1}{2}\beta. \text{ Using Ceva's Theorem} \end{aligned}$$
$$1 = \frac{\sin \angle AND \cdot \sin \angle NBD \cdot \sin \angle BAD}{\sin \angle DNB \cdot \sin \angle DBA \cdot \sin \angle DAN} \text{ or } \frac{\sin \angle DNB}{\sin \angle AND} = \frac{\sin \frac{1}{2}\alpha \cdot \sin(\alpha - \frac{1}{2}\beta)}{\sin \frac{1}{2}\beta \cdot \sin(\beta - \frac{1}{2}\alpha)} \Rightarrow \\ \frac{\sin \angle ENB}{\sin \angle ANE} = \frac{\sin \angle DNB}{\sin \angle AND}$$
Therefore the rays $ND$ and $NE$ coincide since
$$\angle BNA = \angle ANE + \angle ENB = \angle AND + \angle DNB$$
Now let $L$ be the second intersection point of $DE$ and the circumcircle of $\triangle ABC$. We claim that $KP$ passes through $L$. Obviously
$$\angle (AL, LN) = \angle (AB, BN) = \frac{1}{2}(\alpha + \beta).$$
On the other hand,
$$\angle (AP, PB) = \angle (PA, AB) + \angle (AB, BP) = \alpha - \frac{1}{2}\beta + \beta - \frac{1}{2}\alpha = \frac{1}{2}(\alpha + \beta).$$
Hence $\angle (AL, LN) = \angle (AP, PB)$ and the quadrilateral $APDL$ is inscribed. Also we have $\angle (DI, IE) = \frac{1}{2}(\alpha + \beta) = \angle (EP, PB)$. By symmetry $\angle (EK, KD) = \angle (DI, IE)$ and $\angle (KD, DE) = \angle (ED, DI)$, so $\angle (EK, KD) = \angle (EP, PD)$ and $PKED$ is inscribed.
Finally
$$\angle (KP, PE) = \angle (KD, DE) = \angle (ED, DI) = \angle (LD, DA) = \angle (LP, PA).$$
We obtain $\angle (KP, PE) = \angle (LP, PA)$ and since $AP$ and $EP$ coincide we get that the line $KP$ passes through $L$.