Mongolian Mathematical Olympiad

Let $AD$, $CF$ be altitudes dropped from vertices $A$, $C$ respectively. Consequently points $D$, $F$ lie on the circle with diameter $AC$. Let $X$ be intersection point of line $CK$ with line segment $BH$.

Since points $A$, $H$, $K$, $B$ lie on a circle, $\angle KAH = \angle KBH$.

On the other hand points $A$, $K$, $F$, $C$ lie on a circle and furthermore $\angle KAF = \angle KCF$. It yields $\angle KBX = \angle XCB$ and consequently triangles $\triangle KBX$, $\triangle BCX$ are similar. From here we conclude that
$$\frac{KX}{BX} = \frac{BX}{CX},$$
thus $BX^2 = KX \cdot CX$.

On the other hand, points $A$, $H$, $K$, $B$ lie on a circle and consequently $\angle BAK = \angle BHK$. Since points $A$, $D$, $K$, $C$ lie on a circle, $\angle KAD = \angle KCD$.
From here we conclude that $\angle XHK = \angle XCH$, thus triangles $\triangle XHK$, $\triangle XCH$ are similar. Consequently we get
$$\frac{XK}{XH} = \frac{XH}{XC}$$
and $XH^2 = XK \cdot XC$. Combining received results, we conclude
$$BX^2 = KX \cdot CX = XH^2 \text{ and } BX = XH.$$