Iranian Mathematical Olympiad

$$\frac{BD}{CD} \cdot \frac{\sin(\angle A - \angle ABD)}{\sin(\angle A - \angle ACD)} = \frac{\cos \angle B}{\cos \angle C} \quad (\heartsuit)$$
Let $R$ and $Q$ be the intersections of the perpendicular bisectors of $AC$ and $AB$ with $AB$ and $AC$, respectively. It is well-known that $CR$ and $BQ$ meet on $OH$. Let $S$ be the concurrency point of $CR$ and $BQ$. It is obvious that $\angle A - \angle ACD = 180^\circ - \angle DCS$ and $\angle A - \angle ABD = \angle SBD$. So by law of sines
$$\frac{\sin \angle DSB}{\sin \angle DSC} = \frac{BD}{CD} \cdot \frac{\sin \angle DBS}{\sin \angle DCS} = \frac{BD}{CD} \cdot \frac{\sin(\angle A - \angle ABD)}{\sin(\angle A - \angle ACD)} \quad (1)$$

and
$$\frac{\sin \angle OSB}{\sin \angle OSC} = \frac{BO}{CO} \cdot \frac{\sin \angle OBS}{\sin \angle OCS} = \frac{\cos \angle B}{\cos \angle C}. \quad (2)$$
From (1) and (2), $(\heartsuit)$ follows.

Suppose that the circumcircle of triangle $BXE$ intersects the lines $AB$ and $EF$ again at $L$ and $U$, and the circumcircle of triangle $CXF$ intersects the lines $AC$ and $EF$ again at $K$ and $V$. Let $AH$ and $AD$ intersect $EF$ and $BC$ at $J$ and $G$, respectively. We just need to show that $\frac{JE}{JV} = \frac{JF}{JU}$ that is equivalent to $\frac{JE}{EV} = \frac{JF}{FU}$. Notice that
$$\angle KXE = \angle KXF - \angle EXF = (180^\circ - \angle ACD) - (180^\circ - \angle A) = \angle A - \angle ACD.$$
Now by law of sines
$$\frac{EK}{EX} = \frac{\sin \angle KXE}{\sin \angle XKE} = \frac{\sin(\angle A - \angle ACD)}{\sin \angle XFC}.$$
Similarly we have
$$\frac{FL}{FX} = \frac{\sin(\angle A - \angle ABD)}{\sin \angle XEB}.$$

Therefore,
$$\begin{align*} \frac{EK}{FL} &= \frac{EX}{FX} \cdot \frac{\sin \angle XEB}{\sin \angle XFC} \cdot \frac{\sin(\angle A - \angle ACD)}{\sin(\angle A - \angle ABD)} \\ &= \frac{\sin \angle EDA}{\sin \angle FDA} \cdot \frac{\sin(\angle A - \angle ACD)}{\sin(\angle A - \angle ABD)} \tag{3} \\ &= \frac{BG}{CG} \cdot \frac{CD}{BD} \cdot \frac{\sin(\angle A - \angle ACD)}{\sin(\angle A - \angle ABD)} \stackrel{(\heartsuit)}{=} \frac{BG}{CG} \cdot \frac{\cos \angle C}{\cos \angle B}, \end{align*}$$
On the other hand by Ceva's theorem
$$\frac{EK}{FL} = \frac{\frac{EV \cdot EF}{CE}}{\frac{FU \cdot EF}{BF}} = \frac{EV}{FU} \cdot \frac{BG}{CG} \cdot \frac{AF}{AE} \\ \stackrel{(3)}{\implies} \frac{EV}{FU} = \frac{AE}{AF} \cdot \frac{\cos \angle C}{\cos \angle B} = \frac{JE}{JF},$$
as desired.
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