Subfield: Geometry


Question: Let $A B C D$ be a quadrilateral inscribed in a circle $\Omega$. Let the tangent to $\Omega$ at $D$ intersect the rays $B A$ and $B C$ at points $E$ and $F$, respectively. A point $T$ is chosen inside the triangle $A B C$ so that $T E \| C D$ and $T F \| A D$. Let $K \neq D$ be a point on the segment $D F$ such that $T D=T K$. Prove that the lines $A C, D T$ and $B K$ intersect at one point.

Solution:
Let the segments $T E$ and $T F$ cross $A C$ at $P$ and $Q$, respectively. Since $P E \| C D$ and $E D$ is tangent to the circumcircle of $A B C D$, we have

$$
\angle E P A=\angle D C A=\angle E D A
$$

and so the points $A, P, D$, and $E$ lie on some circle $\alpha$. Similarly, the points $C, Q, D$, and $F$ lie on some circle $\gamma$.

We now want to prove that the line $D T$ is tangent to both $\alpha$ and $\gamma$ at $D$. Indeed, since $\angle F C D+\angle E A D=180^{\circ}$, the circles $\alpha$ and $\gamma$ are tangent to each other at $D$. To prove that $T$ lies on their common tangent line at $D$ (i.e., on their radical axis), it suffices to check that $T P \cdot T E=T Q \cdot T F$, or that the quadrilateral $P E F Q$ is cyclic. This fact follows from

$$
\angle Q F E=\angle A D E=\angle A P E .
$$

Since $T D=T K$, we have $\angle T K D=\angle T D K$. Next, as $T D$ and $D E$ are tangent to $\alpha$ and $\Omega$, respectively, we obtain

$$
\angle T K D=\angle T D K=\angle E A D=\angle B D E
$$

which implies $T K \| B D$.

Next we prove that the five points $T, P, Q, D$, and $K$ lie on some circle $\tau$. Indeed, since $T D$ is tangent to the circle $\alpha$ we have

$$
\angle E P D=\angle T D F=\angle T K D
$$

which means that the point $P$ lies on the circle (TDK). Similarly, we have $Q \in(T D K)$.

Finally, we prove that $P K \| B C$. Indeed, using the circles $\tau$ and $\gamma$ we conclude that

$$
\angle P K D=\angle P Q D=\angle D F C
$$

which means that $P K \| B C$.

Triangles $T P K$ and $D C B$ have pairwise parallel sides, which implies the fact that $T D, P C$ and $K B$ are concurrent, as desired.

<img_4033>
Consider the spiral similarity $\phi$ centred at $D$ which maps $B$ to $F$. Recall that for any two points $X$ and $Y$, the triangles $D X \phi(X)$ and $D Y \phi(Y)$ are similar.

Define $T^{\prime}=\phi(E)$. Then

$$
\angle C D F=\angle F B D=\angle \phi(B) B D=\angle \phi(E) E D=\angle T^{\prime} E D,
$$

so $C D \| T^{\prime} E$. Using the fact that $D E$ is tangent to $(A B D)$ and then applying $\phi$ we infer

$$
\angle A D E=\angle A B D=\angle T^{\prime} F D
$$

so $A D \| T^{\prime} F$; hence $T^{\prime}$ coincides with $T$. Therefore,

$$
\angle B D E=\angle F D T=\angle D K T
$$

whence $T K \| B D$.

Let $B K \cap T D=X, A C \cap T D=Y$, and $A C \cap T F=Q$. Notice that $T K \| B D$ implies

$$
\frac{T X}{X D}=\frac{T K}{B D}=\frac{T D}{B D}
$$

So we wish to prove that $\frac{T Y}{Y D}$ is equal to the same ratio.

We first show that $\phi(A)=Q$. Indeed,

$$
\angle D A \phi(A)=\angle D B F=\angle D A C
$$

and so $\phi(A) \in A C$. Together with $\phi(A) \in \phi(E B)=T F$ this yields $\phi(A)=Q$. It follows that

$$
\frac{TQ}{AE}=\frac{TD}{ED}
$$
<img_3959>



Now, since $T F \| A D$ and $\triangle E A D \sim \triangle E D B$, we have

$$
\frac{T Y}{Y D}=\frac{T Q}{A D}=\frac{T Q}{A E} \cdot \frac{A E}{A D}=\frac{T D}{E D} \cdot \frac{E D}{B D}=\frac{T D}{B D}
$$

which completes the proof.