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 $△KBX$, $△BCX$ are similar. From here we conclude that $$\frac{KX}{BX}=\frac{BX}{CX},$$ thus $B{X}^2=KX\cdot CX$.



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 $△XHK$, $△XCH$ are similar. Consequently we get $$\frac{XK}{XH}=\frac{XH}{XC}$$ and $X{H}^2=XK\cdot XC$. Combining received results, we conclude $$B{X}^2=KX\cdot CX=X{H}^2\text{ and }BX=XH.$$