Balti Tee 2020 · Ülesanne 13

Note that $XH \perp BY \perp AC$ and $YH \perp BX \perp AD$. Therefore $XH \parallel AC$ and $YH \parallel AD$. It follows that

$$\frac{AD}{AX} = \frac{CD}{CH} = \frac{CA}{CY} \implies \frac{AD}{CA} = \frac{AX}{CY} = \frac{AB \sin \angle XAB}{CB \sin \angle YCB}.$$

Since $\ell$ is tangent to $\omega$, we have $\angle XAB = \angle YCB$. Thus the sines in the equality above cancel out and we obtain

$$\frac{AD}{CA} = \frac{AB}{CB}.$$

This, along with $\angle DAB = \angle ACB$, proves that $\triangle DAB \sim \triangle ACB$ by SAS. Therefore $\angle CBA = \angle ABD$. This shows that $BA$ bisects angle $CBD$.

Let $D'$ be a point on $\ell$ such that $\angle CBA = \angle ABD'$. Let $Z$ and $T$ be projections of $A$ onto $BC$ and $BD'$, respectively. Note that the circle with diameter $AB$ passes through $X, Y, Z, T$. By Pascal's theorem for hexagon $AXZBTY$, points $D', C$, and $H' := XZ \cap TY$ are collinear. We have

$$\angle XZB = \angle XAB = \angle CAB = 90^\circ - \angle ZBY$$

which shows that $XZ \perp BY$. By definition of $D'$,

$$\angle BD'A = 180^\circ - \angle D'AB - \angle ABD' = 180^\circ - \angle ACB - \angle CBA = \angle BAC.$$

Therefore

$$\angle D'AT = 90^\circ - \angle TD'A = 90^\circ - \angle BAC,$$

hence

$$\angle XYT + \angle BXY = \angle XAT + \angle BAY = 90^\circ - \angle BAC + \angle BAC = 90^\circ.$$

This shows that $YT \perp BX$. Since $XZ \perp BY$ and $YT \perp BX$, it follows that $H'$ is the orthocenter of $BXY$, i.e. $H' = H$. Since $D', C, H'$ are collinear and $D'$ lies on $\ell$, it follows that $D' = D$. Therefore $\angle ABD = \angle CBA$ and we are done.