Korean Mathematical Olympiad Final Round

Observation 1. $A$, $B$, $C$, $Z$ are cyclic.
$$\angle BAD = \angle BXD = \angle ZBC \text{ and } \angle DAC = \angle DYC = \angle BCZ.$$
Hence $\angle BAC = \angle BAD + \angle BCZ = 180^\circ - \angle BZC$, that is, $\angle BAC + \angle BZC = 180^\circ$.

Observation 2. Both $A$, $E$, $Z$, $Y$ and $A$, $X$, $Z$, $F$ are cyclic.
By Observation 1, we have
$$\angle AEY = \angle ABC = \angle AZC = \angle AZY.$$
Similarly, $\angle AFX = \angle AZX$.

Lemma 1. $ZE = ZF$ if and only if $\frac{BD}{CD} = \frac{AB}{AC}$.

Proof. We can easily see that $\triangle AEZ$ is similar to $\triangle ADC$, and symmetrically, $\triangle AFZ$ is similar to $\triangle ADB$. Then we have
$$ZE = \frac{AE \cdot CD}{AD} \quad \text{and} \quad ZF = \frac{AF \cdot BD}{AD}.$$
Therefore, we have $ZE = ZF \Leftrightarrow \frac{BD}{CD} = \frac{AE}{AF} = \frac{AB}{AC}$.

Now let $Q$, $R$, $S$ be the feet of perpendiculars from $A$, $B$, $C$ to $BC$, $CA$, $AB$, respectively. Then $A$, $S$, $D$, $H$, $F$ are cyclic. Since $AD \cdot AP = AH \cdot AQ = AS \cdot AB$, $S$, $D$, $P$, $B$ are cyclic.
Similarly, $R$, $D$, $P$, $C$ are cyclic.
Since $AS : AP = AD : AB$, $\triangle ASP$ is similar to $\triangle ADB$.
Now we have
$$(1) \qquad PS = BD \frac{AP}{AB}, \quad RP = CD \frac{AP}{AC}.$$

Lemma 2. Let $P$ be a point on $BC$. Then
$$BP = PC \Leftrightarrow PR = PS.$$
Proof. Suppose $P$ is a midpoint of $BC$. Then $P$, $Q$, $R$, $S$ are on the nine-point circle, so we have
$$\angle PSR = \angle RQC = \angle A = \angle SQB = \angle PRS.$$
That is, $PR = PS$. Hence $P$ is the intersection of the perpendicular bisector of $SR$ and $BC$, so it is unique. Thus $PR = PS$ implies that $P$ is a midpoint of $BC$. $\square$

Now we are ready to prove that $BP = PC \Leftrightarrow ZE = ZF$.

Suppose $BP = PC$, then by Lemma 2, $PR = PS$. Then by (1), we have $\frac{BD}{CD} = \frac{AB}{AC}$. Then by Lemma 1, $ZE = ZF$.

Conversely, assume that $ZE = ZF$, then by Lemma 1, $\frac{BD}{CD} = \frac{AB}{AC}$. Then by (1), we have $PS = PR$. By Lemma 2, $BP = PC$. $\square$