A problem with two interlinked right triangles is driving me crazy

Question

This is a problem that is driving me crazy:

$F$ is the center of the circle.

$A\hat{B}C$ is a right angle.

$A\hat{B}F$ is a right angle.

$D\hat{B}F$ is a right angle.

$C\hat{E}F$ is a right angle.

The length $\overline{BF}$ is known.

The angle $\alpha=B\hat{A}C$ is equal to the angle $B\hat{F}D$ and it is known to be different from zero.

Given $\alpha$ and $\overline{BF}$ is it possible to compute $\overline{EF}$?

The only thing I can write, apart from a bunch of equations deriving from the Pythagorean theorem like $\overline{CF}^2=\overline{CE}^2+\overline{EF}^2$, is

$\overline{DF}=\frac{\overline{BF}}{\cos{\alpha}}$

but I need $\overline{EF}\lt\overline{DF}$ and not $\overline{DF}$.

I really can't see the solution...

Assuming some more info: given $\alpha$ and $\overline{BF}$ and $\overline{AB}$ is it possible to compute $\overline{EF}$?

Answer 1

The data $\alpha$ and $s:=|BF|$ alone do not determine $t:=|EF|$. In order to see this, let $M$ be the midpoint of $AF$. Then $M$ is the center of a Thales circle over the diameter $AF$ and containing $B$, $E$ on its periphery. Since we are not told the angle $\beta:=\angle(BFA)$, or the diameter $|AF|$ of this circle, the knowledge of $s$ ad $\alpha$ is insufficient to determine $t$.

If $|AB|=:a$ is given as well we can compute $\beta=\arctan{a\over s}$ and then $d:=|AF|=\sqrt{a^2+s^2}$. Finally $t=|EF|=d\cos(\alpha+\beta)$.

(The drawing was done using Canvas Draw.)

Answer 2

I have simplified your drawing a bit - so, the angle at B is right, and the angle a E is also right. If you choose point p so that the angle there is right, then the distance from F to p is $|BF| \cdot cos(v)$, and the distance from D to p is $|BF| \cdot sin(v)$, since the angl DBp must be v: the sum of angles in a triangle is 180, BDp shares an angle at D, and both have a right angle. So there you are: $|DF| = |BF| \cdot (sin(v) + cos(v))$