Consider a cyclic quadrilateral $ABCD$ where
$|DA|\lt |AB|=|BC|\lt |CD|$, $E:|BE|\bot |AC|, E\in |CD|$ , $F:|EF|\text{ }||\text{ }|BC|,F\in |AB|$.
The task is to prove that |FB|=|FD|, that is, that $\Delta FDB$ is isosceles.
I have been able to prove some statements about the quadrilateral, like tha fact that the triangles formed by the diagonals opposite each other have to be similar, but I am struggling to prove anything of use regarding $\Delta ABC$ and $\Delta FDB$.
Thank you for any help.
On
Draw $AE$ and $FD$.
Observe that, $AFED$ is a cyclic quadrilateral.
Also, since $\triangle ABC$ is isosceles, $BE$ is the perpendicular bisector of $AC$.
Thus, $\angle FDB=\angle FDE-\angle BDE=\angle FDE-\angle BDC=\angle FAE-\angle BAC=\angle FAE-\angle FAC=\angle CAE=\angle ACE=\angle ACD=\angle ABD=\angle FBD$.
Therefore, $\triangle FDB$ is isosceles.
Edit : The reason why $AFED$ is cyclic is because since $EF$ is parallel to $BC$, $\angle AFE=\angle ABC=180-\angle ADC=180-\angle ADE$ and hence $AFED$ is cyclic.[It's opposite angles are supplementary.] ( Credit : User @V.S.e.H.)
Let $\mathcal{C}$ be the circle through $A, B, C, D$ and denote by $G$ the point of intersections of the diagonals $AC$ and $BD$. Furthermore, call $\mathcal{X}$ and $\mathcal{Y}$ the circles through $AGD$ and $DGC$, respectively.
Because we have the equality of angles $\angle BAC = \angle BCA = \angle BDA$, where we have used that the triangle $ABC$ is isosceles and carried an angle through the circle, we deduce that $\triangle BGA \sim \triangle BAD$, which implies $BA^2 = BG\cdot BD$. Hence, $\mathcal{X}$ is tangent to $BA$ at $A$. In analogous fashion we deduce that $\mathcal{Y}$ is tangent to $BC$ at $C$.
If we call $M$ the other point of intersection of $BE$ with circle $\mathcal{C}$ we know, since $BE\perp AC$ and $ABC$ is isosceles, that that $BM$ is a diameter of $\mathcal{C}$ and hence $MC\perp BC$ and $MA\perp BA$. Because of the tangencies proves above the center of $\mathcal{X}$ has to lie on $AM$ and that of $\mathcal{Y}$ on $MC$.
Call $X$ the intersection of $AM$ and the parallel line through $E$ to $MC$. We have then $$\angle DAM = \angle DCM = \angle DEX.$$ Hence we deduce $DAEX$ is a cyclic quadrilateral.
Now call $L$ the intersection of $DX$ with $MC$. We have that
$$\begin{align*} \angle LDC &= \angle XDE\\ &= \angle XAE &\;\mbox{(by previous cyclic quadrilateral)}\\ &= \angle XAC - \angle EAC\\ &= \angle MAC - \angle EAC\\ &= \angle MCA - \angle ECA &\mbox{(since BM is perp. bisector of AC)}\\ &= \angle MCE \end{align*}$$
Hence, $LD = LC$ and so $L$ lies in the perpendicular bisector of $DC$. But the center of $\mathcal{Y}$, which also lies in that perpendicular bisector, since $DC$ is a chord, has to be in $MC$ since this is orthogonal to the tangent $BC$ at the tangency point $C$. We conclude $L$ is the center of $\mathcal{Y}$.
Now, $\angle DAF = \angle DAB = 180 - \angle BCD = 180 - \angle FED$. Hence, $DAFE$ is a cyclic quadrilateral, but $DAEX$ was also a cyclic quadrilateral, and since these circles share the points $D, A, E$, we conclude $DAFEX$ is a cyclic pentagon.
Now, $\angle FDL = \angle FDX = 180 - \angle XEF$, using the above cyclic pentagon and, furthermore, by contruction $XE || MC$ and $FE || BC$ so that $\angle XEF = \angle MCB = 90$, since $MC$ lies on a diameter of $\mathcal{Y}$ and $BC$ a tangent. We deduce $\angle FDL = 90$.
We have proven above that $L$ is the center of $\mathcal{Y}$, and since $D$ is on $\mathcal{Y}$ then $DF$ is a tangent to this circle since we just proved $\angle FDL = 90$. Hence, transporting angles, first on $\mathcal{Y}$, using this tangent, and then on $\mathcal{C}$ we conclude $$\angle FDG = \angle DCG = \angle DCA = \angle DBA = \angle DBF,$$ that is, $\triangle FDB$ is isosceles, as desired.
Here is a drawing: