It approximately looks like the following picture
The figure may be rotated at any angle.
I know the coordinates of points $A$, $B$, $C$, $D$ and the length of $BF$.
$\angle ABD$ and $\angle CBD$ are equal ($AD = CD$ and $AB = CB$).
$\angle EAB = \angle EFB = \frac{\pi}{2}$ (right angles).
I need to find the coordinates of $E$ and $G$.
Construct lines AD and СD by given coordinates of points A, С and D. Let it be in parametric form. AD: { x = A.x + (D.x - A.x)*t, y = A.y + (D.y - A.y)*t } BD: { x = С.x + (D.x - С.x)*t, y = C.y + (D.y - С.y)*t } In our definitions 0 <= t <= 1, but actually it doesn't matter in future.
Find coordinates of point F. Let vector V := D - B = {D.x - B.x, D.y - B.y}. Then we do following steps: V = V / |V|, V = V*|BF|. Here |V| = sqrt(V.x*V.x + V.y*V.y) - the length of the vector V, and |BF| - the known length of the segment BF. Now F can be obtained clearly: F = {B.x + V.x, B.y + V.y}.
Construct line EG: we know, that this line contains point F and this line is orthogonal to BD. Hence this line has direction vector orthogonal to vector V. Construct this vector: let W := {-V.y, V.x}. So, our line has the following parametric form: EG: { x = F.x + W.x*u, y = F.y + W.y*u }
Finally, find intersection of lines AD and EG, CD and EG. For this we have to solve two system of two equations with respect to t and u. For AD and EG it is: A.x + (D.x - A.x)*t = F.x + W.x*u, A.y + (D.y - A.y)*t = F.y + W.y*u. In your case it has the only solution, where t = (A.y*W.x - A.x*W.y + F.x*W.y - F.y*W.x) / (A.y*W.x - A.x*W.y + D.x*W.y - D.y*W.x); Just paste this value of t into equation of line AD (p.1) - and you will get your point E. The same way you can obtain point G as intersection of lines CD and EG. For it t = (C.y*W.x - C.x*W.y + F.x*W.y - F.y*W.x) / (C.y*W.x - C.x*W.y + D.x*W.y - D.y*W.x);
Pasting this value into equation of line BD, you will get you point G.