Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral with $\angle B=\angle D=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $B D$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle $S C T$ and

$$
\angle S H C-\angle B S C=90^{\circ}, \quad \angle T H C-\angle D T C=90^{\circ} .
$$

Prove that the circumcircle of triangle $S H T$ is tangent to the line $B D$.

Solution:
Let the line passing through $C$ and perpendicular to the line $S C$ intersect the line $A B$ at $Q$ (see Figure 1). Then

$$
\angle S Q C=90^{\circ}-\angle B S C=180^{\circ}-\angle S H C
$$

which implies that the points $C, H, S$, and $Q$ lie on a common circle. Moreover, since $S Q$ is a diameter of this circle, we infer that the circumcentre $K$ of triangle $S H C$ lies on the line $A B$. Similarly, we prove that the circumcentre $L$ of triangle $C H T$ lies on the line $A D$.

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Figure 1

In order to prove that the circumcircle of triangle $S H T$ is tangent to $B D$, it suffices to show that the perpendicular bisectors of $H S$ and $H T$ intersect on the line $A H$. However, these two perpendicular bisectors coincide with the angle bisectors of angles $A K H$ and $A L H$. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that

$$
\frac{A K}{K H}=\frac{A L}{L H}\tag{1}
$$

We present two proofs of this equality.

First proof. Let the lines $K L$ and $H C$ intersect at $M$ (see Figure 2). Since $K H=K C$ and $L H=L C$, the points $H$ and $C$ are symmetric to each other with respect to the line $K L$. Therefore $M$ is the midpoint of $H C$. Denote by $O$ the circumcentre of quadrilateral $A B C D$. Then $O$ is the midpoint of $A C$. Therefore we have $O M \| A H$ and hence $O M \perp B D$. This together with the equality $O B=O D$ implies that $O M$ is the perpendicular bisector of $B D$ and therefore $B M=D M$.

Since $C M \perp K L$, the points $B, C, M$, and $K$ lie on a common circle with diameter $K C$. Similarly, the points $L, C, M$, and $D$ lie on a circle with diameter $L C$. Thus, using the sine law, we obtain

$$
\frac{A K}{A L}=\frac{\sin \angle A L K}{\sin \angle A K L}=\frac{D M}{C L} \cdot \frac{C K}{B M}=\frac{C K}{C L}=\frac{K H}{L H}
$$



which finishes the proof of (1).

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Figure 2

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Figure 3

Second proof. If the points $A, H$, and $C$ are collinear, then $A K=A L$ and $K H=L H$, so the equality (1) follows. Assume therefore that the points $A, H$, and $C$ do not lie in a line and consider the circle $\omega$ passing through them (see Figure 3). Since the quadrilateral $A B C D$ is cyclic,

$$
\angle B A C=\angle B D C=90^{\circ}-\angle A D H=\angle H A D .
$$

Let $N \neq A$ be the intersection point of the circle $\omega$ and the angle bisector of $\angle C A H$. Then $A N$ is also the angle bisector of $\angle B A D$. Since $H$ and $C$ are symmetric to each other with respect to the line $K L$ and $H N=N C$, it follows that both $N$ and the centre of $\omega$ lie on the line $K L$. This means that the circle $\omega$ is an Apollonius circle of the points $K$ and $L$. This immediately yields (1).