Subfield: Geometry


Question: Let $A B C$ be a triangle with $\angle C=90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $C B H$ so that $C H$ bisects $A D$. Let $P$ be the intersection point of the lines $B D$ and $C H$. Let $\omega$ be the semicircle with diameter $B D$ that meets the segment $C B$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $C Q$ and $A D$ meet on $\omega$.

Solution:
Let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 1). Since $P H \| D K$, we have

$$
\frac{P D}{P B}=\frac{H K}{H B}=\frac{A H}{H B}
\tag{1}
$$

Let $L$ be the projection of $Q$ onto $D B$. Since $P Q$ is tangent to $\omega$ and $\angle D Q B=\angle B L Q=$ $90^{\circ}$, we have $\angle P Q D=\angle Q B P=\angle D Q L$. Therefore, $Q D$ and $Q B$ are respectively the internal and the external bisectors of $\angle P Q L$. By the angle bisector theorem, we obtain

$$
\frac{P D}{D L}=\frac{P Q}{Q L}=\frac{P B}{B L}
\tag{2}
$$

The relations (1) and (2) yield $\frac{A H}{H B}=\frac{P D}{P B}=\frac{D L}{L B}$. So, the spiral similarity $\tau$ centered at $B$ and sending $A$ to $D$ maps $H$ to $L$. Moreover, $\tau$ sends the semicircle with diameter $A B$ passing through $C$ to $\omega$. Due to $C H \perp A B$ and $Q L \perp D B$, it follows that $\tau(C)=Q$.

Hence, the triangles $A B D$ and $C B Q$ are similar, so $\angle A D B=\angle C Q B$. This means that the lines $A D$ and $C Q$ meet at some point $T$, and this point satisfies $\angle B D T=\angle B Q T$. Therefore, $T$ lies on $\omega$, as needed.

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Figure 1

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Figure 2
Let $\Gamma$ be the circumcircle of $A B C$, and let $A D$ meet $\omega$ at $T$. Then $\angle A T B=$ $\angle A C B=90^{\circ}$, so $T$ lies on $\Gamma$ as well. As in the previous solution, let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 2).

Our goal now is to prove that the points $C, Q$, and $T$ are collinear. Let $C T$ meet $\omega$ again at $Q^{\prime}$. Then, it suffices to show that $P Q^{\prime}$ is tangent to $\omega$, or that $\angle P Q^{\prime} D=\angle Q^{\prime} B D$.

Since the quadrilateral $B D Q^{\prime} T$ is cyclic and the triangles $A H C$ and $K H C$ are congruent, we have $\angle Q^{\prime} B D=\angle Q^{\prime} T D=\angle C T A=\angle C B A=\angle A C H=\angle H C K$. Hence, the right triangles $C H K$ and $B Q^{\prime} D$ are similar. This implies that $\frac{H K}{C K}=\frac{Q^{\prime} D}{B D}$, and thus $H K \cdot B D=C K \cdot Q^{\prime} D$.

Notice that $P H \| D K$; therefore, we have $\frac{P D}{B D}=\frac{H K}{B K}$, and so $P D \cdot B K=H K \cdot B D$. Consequently, $P D \cdot B K=H K \cdot B D=C K \cdot Q^{\prime} D$, which yields $\frac{P D}{Q^{\prime} D}=\frac{C K}{B K}$.

Since $\angle C K A=\angle K A C=\angle B D Q^{\prime}$, the triangles $C K B$ and $P D Q^{\prime}$ are similar, so $\angle P Q^{\prime} D=$ $\angle C B A=\angle Q^{\prime} B D$, as required.
Introduce the points $T$ and $Q^{\prime}$ as in the previous solution. Note that $T$ lies on the circumcircle of $A B C$. Here we present yet another proof that $P Q^{\prime}$ is tangent to $\omega$.

Let $\Omega$ be the circle completing the semicircle $\omega$. Construct a point $F$ symmetric to $C$ with respect to $A B$. Let $S \neq T$ be the second intersection point of $F T$ and $\Omega$ (see Figure 4).

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Figure 4

Since $A C=A F$, we have $\angle D K C=\angle H C K=\angle C B A=\angle C T A=\angle D T S=180^{\circ}-$ $\angle S K D$. Thus, the points $C, K$, and $S$ are collinear. Notice also that $\angle Q^{\prime} K D=\angle Q^{\prime} T D=$ $\angle H C K=\angle K F H=180^{\circ}-\angle D K F$. This implies that the points $F, K$, and $Q^{\prime}$ are collinear.

Applying PASCAL's theorem to the degenerate hexagon $K Q^{\prime} Q^{\prime} T S S$, we get that the tangents to $\Omega$ passing through $Q^{\prime}$ and $S$ intersect on $C F$. The relation $\angle Q^{\prime} T D=\angle D T S$ yields that $Q^{\prime}$ and $S$ are symmetric with respect to $B D$. Therefore, the two tangents also intersect on $B D$. Thus, the two tangents pass through $P$. Hence, $P Q^{\prime}$ is tangent to $\omega$, as needed.