Problem 1926

Let $O$ be the circumcentre of $\odot(ABC)$. We proceed in several steps.

Step 1: Point $O$ lies on $\odot(ABD)$ and $\odot(ACE)$.
Proof. Note that since $|AD| = |DC|$, we have $\angle CDA = 180^\circ - 2\angle ACB$, therefore $\angle ADB = 2\angle ACB = \angle AOB$, which means that quadrilateral $AODB$ is cyclic. Similarly, we can prove that quadrilateral $AOEC$ is cyclic. $\square$

Step 2: Points $Y$, $A$, $Z$ are collinear. Moreover, $YZ$ is tangent to $\odot(ABC)$ at point $A$.
Proof. Note that since $YB$ is tangent to $\odot(ABC)$, we have that:
$$\angle AOY = \angle YBA = \angle ACB.$$
Since $\angle AOB = 2\angle ACB$, we get that:
$$\angle YAB = \angle YOB = \angle AOB - \angle AOY = \angle ACB.$$
This means that $YA$ is tangent to $\odot(ABC)$. Consequently, $|YA| = |YB|$ implies that $Y$ lies on the perpendicular bisector of $AB$. Similarly, we can prove that $AZ$ is tangent to $\odot(ABC)$ and $Z$ lies on the perpendicular bisector of $AC$. Since the tangent line at a fixed point is unique, we conclude that points $Y$, $A$, $Z$ all lie on the tangent to $\odot(ABC)$ at the point $A$. $\square$

Step 3: Points $Y$, $O$, $E$ and $Z$, $O$, $D$ are collinear. Moreover, $O$ is the orthocentre of $\triangle YZX$.
Proof. The first part follows from the previous result that $Y$ lies on the perpendicular bisector of $AB$, which is $OE$. A similar argument applies to $Z$, $O$, $D$.
Since the radius of a circle is perpendicular to the corresponding tangent, note that:
$$\angle OBY = \angle ODY = \angle YAO = 90^\circ \quad \text{and} \quad \angle OCZ = \angle OEZ = 90^\circ.$$
This gives us that $ZD \perp YX$ and $YE \perp ZX$, implying that $O$ is the orthocentre of $\triangle YZX$. Also note that this immediately gives us that points $A$, $O$, $X$ are collinear, too. $\square$

To finish the problem, note that quadrilateral $YDEZ$ is cyclic from $\angle YDZ = \angle YEZ = 90^\circ$. Therefore:
$$\angle DAO = \angle DYO = \angle OZE = \angle OAE.$$

Since $OX$ is the diameter of $\odot(DEX)$ (note that $\angle ODX = \angle OEX = 90^\circ$) and $OX$ is the bisector of $\angle DAE$, by symmetry we have that $DV$ and $UE$ are parallel lines. It is well-known that the diagonals of an isosceles trapezoid intersect on the perpendicular bisector of their parallel sides. Therefore, lines $UV$, $DE$, $AX$ are concurrent, as desired.