Japan Mathematical Olympiad

If $AB=AC$ holds, the center of a circle going through the points $B$, $C$ and the center of the circle having $AH$ as its diameter both lie on the perpendicular bisector of the side $BC$ of the triangle. Therefore, the line $XY$ and the perpendicular bisector of $BC$ intersect perpendicularly and the point of their intersection coincides with the point $K$. Furthermore, the point $D$ is the mid-point of the line segment $BC$. Therefore, it is clear that $\angle BKD=\angle CKD$ holds.

So, we consider the case where $AB\ne AC$. We may assume that $AB>AC$.

Let $E$ be the point of intersection of the lines $BH$ and $AC$, and $F$ be the point of intersection of the lines $CH$ and $AB$. Since $\angle AEH=\angle AFH=90^{\circ }$, the points $E$, $F$ lie on the circumference of the circle having $AH$ as its diameter. Call this circle ${\Gamma }_1$. Note that the points $X$, $Y$ also lie on ${\Gamma }_1$. Note also that 4 points $B$, $C$, $X$, $Y$ lie on the circumference of a circle, which we denote by ${\Gamma }_2$, and 4 points $B$, $C$, $E$, $F$ lie on the circumference of another circle, which we denote by ${\Gamma }_3$.

Lemma. 3 lines $XY$, $BC$ and $EF$ intersect at 1 point.

Proof. Let $O$ be the point of intersection of the two lines $BC$ and $EF$. Denote by $Y^{\prime }$ and $Y^{\prime \prime }$ the points of intersection of the line $OX$ and the circles ${\Gamma }_1$ and ${\Gamma }_2$, respectively. By the well-known theorem on power of a point with respect to a circle, we get $$OX\cdot O{Y}^{\prime }=OE\cdot OF=OC\cdot OB=OX\cdot O{Y}^{\prime \prime }.$$ Therefore, we have $O{Y}^{\prime }=O{Y}^{\prime \prime }$. Thus, the two points $Y^{\prime }$ and $Y^{\prime \prime }$ must coincide. Since $Y^{\prime }$ lies on the circle ${\Gamma }_1$ and $Y^{\prime \prime }$ lies on the circle ${\Gamma }_2$, we see that $Y^{\prime }=Y^{\prime \prime }=Y$. This shows that the point $O$ lies on the line $XY$ and proves the Lemma.

By Ceva's Theorem we have $$\frac{AF}{FB}\cdot \frac{BD}{DC}\cdot \frac{CE}{EA}=1,$$ and by Menelaus' Theorem we also have $$\frac{AF}{FB}\cdot \frac{BO}{OC}\cdot \frac{CE}{EA}=1.$$ From these two equations, we get $\frac{BD}{DC}=\frac{BO}{OC}$. Let $C^{\prime }$ be the point on the line segment $DO$ satisfying the condition $\angle BKD=\angle C^{\prime }KD$. The line $KD$ is the bisector of the $\angle BK{C}^{\prime }$, and from $\angle DKO=90^{\circ }$ it follows that the line $KO$ is the bisector of $\angle EKO$. Thus we obtain $$\frac{BD}{D{C}^{\prime }}=\frac{KB}{K{C}^{\prime }},\frac{BO}{O{C}^{\prime }}=\frac{KB}{K{C}^{\prime }}$$ Therefore, we conclude that $\frac{BD}{D{C}^{\prime }}=\frac{BO}{O{C}^{\prime }}$. Combining this with $\frac{BD}{DC}=\frac{BO}{OC}$, obtained above, we get $\frac{OC}{DC}=\frac{O{C}^{\prime }}{D{C}^{\prime }}$. Since both $C$ and $C^{\prime }$ lie on the line segment $DO$, we conclude that $C^{\prime }=C$ must hold, and thus we get $\angle BKD=\angle CKD$, which establishes the assertion of the problem.