Find the angle

can have this

Let the position coordinates of $A, D, C$ be $(0,0), \left (\dfrac{b}{2},0\right ), (b, 0)$ respectively. Then those of $B$ are $\left (\dfrac{b}{2}, \dfrac{\sqrt {4a^2-b^2}}{2}\right )$ , where $a=|\overline {BC}|=|\overline {AB}|$ . Solving equations for $\overline {AB}$ and $\overline {DE}$ we get the position coordinates of $E$ as $\left (\dfrac{b^3}{8a^2}, \dfrac{b^2\sqrt {4a^2-b^2}}{8a^2}\right)$ . Let $\overline {DE}$ and $\overline {BF}$ meet at $G$ . Then the position coordinates of $G$ are $\left (\dfrac{b^3+4a^2b}{16a^2}, \dfrac{b^2\sqrt {4a^2-b^2}}{16a^2}\right)$ .

So, the slope of $\overline {BG}$ is $m_1=\dfrac{8a^2-b^2}{b\sqrt {4a^2-b^2}}$ .

Slope of $\overline {CE}$ is $m_2=\dfrac{-b\sqrt {4a^2-b^2}}{8a^2-b^2}$ .

So, $m_1m_2=-1\implies \angle {BFC}=\boxed {90\degree}$ .