Problem 1994

We start our solution with the following lemma.
Lemma. Let $H$ be the orthocenter of the triangle $ABC$ and $E, F$ be the intersections of $BH, CH$ with perpendicular bisectors of $AB$ and $AC$, respectively. If $Q$ is the intersection of tangents from $E$ and $F$ to circumcircle of triangle $HEF$, then $BQ = CQ$.
Proof. Let $L$ and $K$ be the intersections of the perpendicular bisectors of $BC$ with $BH$ and $CH$, respectively. According to the pascal's theorem, it suffices to prove that $KE$ and $LF$ intersect on the circumcircle of $HEF$. In doing so, we should prove that $\angle BEK = \angle CFL$. By symmetry, we have

$\angle EBK = \angle FCL$. If we can show $\frac{BE}{BK} = \frac{CF}{CL}$, then triangles $BEK, CFL$ are similar and we have $\angle BEK = \angle CFL$ as desired.
$$BE = \frac{BN}{\cos \angle EBA} = \frac{AB}{2 \sin \angle A} \\ BK = \frac{CM}{\cos \angle KCB} = \frac{BC}{2 \sin \angle B} \Rightarrow \frac{BE}{BK} = \frac{AB \cdot \sin \angle B}{BC \cdot \sin \angle A},$$
similarly we have $\frac{CF}{CL} = \frac{AC \sin \angle C}{BC \sin \angle A}$, by using law of sines we shall obtain the desired equality.

Now for the original problem, assume that $H$ is the orthocenter of $AXD$ and $E$ and $F$ are the intersections of $AH$ and $BH$ with $CK$ and $CL$, respectively. We have $BE \perp CX$ and $\angle AXD + \angle BXC = 180^\circ$. Therefore,
$$\angle EBX = \angle BXC - 90^\circ = 90^\circ - \angle AXD = \angle EAX.$$
Hence $AEXB$ is cyclic and $AE = AX$. Similarly $CXFD$ is cyclic and $DF = FX$. Let $P$ be the intersection of $BE$ and $CF$, we have
$$\angle AEB = \angle AXB = 180^\circ - \angle DXC = 180^\circ - \angle DFC = \angle HFP.$$
Implying that $HEPF$ is cyclic. Let $Q$ be the intersection of tangents from $E$ and $F$ to the circumcircle of $HEPF$, by Pascal's theorem $Q$ lies on $KL$ and by the above lemma we have $AQ = DQ$. If $Q$ and $O$ are two different points then we have $QO \perp AD$, as desired.

If $Q \equiv O$, then it is clear that $HF \parallel OE$. Therefore $\angle EHF = \angle OEF = \angle EFH$. Similarly $\angle EHF = \angle FEH$, hence the triangle $HEF$ is equilateral and $\angle EHF = 60^\circ$. This implies that $\angle AXD = 60^\circ$ and
$$\angle ABX = 180^\circ - \angle AEX = 2\angle EAX = 60^\circ.$$
Analogously, we have $\angle DCX = 60^\circ$, and
$$\angle ABX + \angle DCX = 120^\circ = 180^\circ - \angle AXD = \angle BXC.$$
But this implies that $AB \parallel CD$, a contradiction. ■