In the figure below, $\angle A B F=\angle F B C=45^{\circ}$ and $A C D E$ is a square. If $A B=2 / 3 \cdot B C$, determine the ratio $\frac{E F}{F D}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-18.jpg?height=586&width=617&top_left_y=518&top_left_x=617)
Solution

Through points $E$ and $D$, the lines parallel to sides $B C$ and $A B$ of triangle $\triangle A B C$ determine, together with the extensions of these sides, points $G, H$, and $I$, as indicated in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-18.jpg?height=539&width=537&top_left_y=1478&top_left_x=657)

Given that $\angle A B C=90^{\circ}$, it follows that $G H I B$ is a rectangle. Since $\angle B A C + \angle A C B = 90^{\circ}$ and $\angle A C D = 90^{\circ}$, it follows that $\angle D C I = \angle B A C$. Thus, triangle $\triangle D C I$ has the same angles as triangle $\triangle A B C$ and $A C = C D$. By the $A L A$ congruence case, these two triangles are congruent. Similarly, we can show that $\triangle A B C \equiv \triangle G A E \equiv \triangle E H D$. Consequently, $G H = G E + E H = A B + B C = H D + D I$ and, therefore, $G H I B$ is a square. This implies that $\angle G H B = \angle B H I = 45^{\circ}$. By the Angle Bisector Theorem applied to triangle $\triangle E H D$, we have:

$$
\begin{aligned}
\frac{E F}{F D} & = \frac{E H}{H D} \\
& = \frac{B C}{A B} \\
& = \frac{3}{2}
\end{aligned}
$$