Baltic Way Shortlist, 2023

Let $K$ be the orthogonal projection of $B$ onto the angle bisector of $\angle BAC$. Now by the Iran lemma and its analogue for the excircle (both easy angle chases), $K$ lies on lines $DE$ and $PQ$. Now let $L$ be the orthogonal projection of $K$ onto line $AB$. Our main claim is that $L$ lies on both $XY$ and $WY$ and thus $X$, $Y$, $W$ are colinear. This also suffices since then a symmetric claim with respect to $C$ gives $X$, $Y$, $Z$ colinear. We use directed angles.

Claim: $L$ lies on $XY$.
Proof: Clearly $KLFY$ is cyclic. Also letting $I$ be the incenter of $\triangle ABC$, we see that $BFIK$ is cyclic. Now
$$\angle XYD = \angle XFE = \angle CDE$$
which is quite easy to chase to be
$$\angle BIK = \angle BFK = \angle LFK = \angle LYK = \angle LYD.$$

Claim: $L$ lies on $WY$.
Proof: First note that $\angle QKE = 90^\circ$ as
$$\angle KEQ + \angle EQK = \angle DEC + \angle CQP = 90^\circ$$
and since $FR$ and $EQ$ are symmetric with respect to $AK$, we also have $\angle FKR = 90^\circ$. Now as $KLRW$ is cyclic, we have
$$\angle WLR = \angle WKR = \angle QKR$$
and by the perpendicularities, we have that this is just
$$\angle EKF = \angle YKF = \angle YLF.$$

So $L$ lies on both $XY$ and $WY$, and thus $X$, $Y$, $W$ are collinear. By symmetry, $X$, $Y$, $Z$ are also collinear, so $X$, $Y$, $Z$, $W$ are collinear.

First we show that $XY$ and $BC$ are parallel. Indeed, since $EFXY$ is cyclic and $BC$ is tangent to the incircle at $D$, we have
$$\angle DXY = \angle FED = \angle FDB = \angle XDB.$$
Similarly, $ZW$ and $BC$ are parallel. Moreover, the points $X$, $Y$, $Z$ and $W$ all lie on the same side of $BC$ because, for example, $X$ lies in the interior of segment $DF$ (since $\angle EDF < 90^\circ$), and similarly for the other points. Therefore it suffices to show that the distance from $BC$ to $XY$ and $ZW$ is the same.

We will use the following Lemma:
Lemma. Let $\triangle ABC$ be a non-rectangular triangle with altitudes $BU$ and $CV$. Then $d(A, UV) = d(A, BC) \cdot |\cos \angle A|$, where $d(P, \ell)$ denotes the distance from point $P$ to line $\ell$.
Proof. Let $\phi$ be the homothety with center $A$ and factor $\cos \angle A$, followed by reflection across the internal angle bisector of $\angle A$. Then $\phi$ maps $B$ to $U$ and $C$ to $V$. Hence $\triangle ABC$ and $\triangle AUV$ are similar with factor $|\cos \angle A|$, from which the Lemma follows.

After applying the Lemma to $\triangle DEF$ and $\triangle PQR$ and noting that
$$\cos \angle RPQ = \cos(180^\circ - \angle EDF) = -\cos \angle EDF,$$
it remains to show that $d(D, EF) = d(P, QR)$.
However, since $EF$ and $RQ$ are parallel and $|BD| = |PC|$, this is equivalent to showing that $d(B, EF) = d(C, QR)$. But this can be done as follows:
$$\begin{align*} d(B, EF) &= |BF| \cdot \sin \angle BFE = |BD| \cdot \sin \angle EFA \\ &= |PC| \cdot \sin \angle AEF = |CQ| \cdot \sin \angle AQR \\ &= d(C, QR). \end{align*}$$

Therefore, the lines $XY$ and $ZW$ are parallel and equidistant from $BC$, so the points $X$, $Y$, $Z$, $W$ are collinear.