1. You are dealing with 2 right triangles: and .
So the point A is located on a half circle (circle of Thales) over BD and point A is located on a half circle (circle of Thales) over CE.
2. You certainely have made an exact drawing(?). See attachment.
a) The point A is the intersection of the green circle and the circle of Thales over BD. The center of the green circle has the coordinates with the radius (why?)
b) Draw the line AC. Construct a right angle in A on AC. The leg of this right angle intersects the prolonged line BD in E.
3. I've done the construction in a coordinate system so you can read the length of x.
4. The line segment BD is divided harmonically by C (inner point) and E (outer point). According to the harmonic partition you'll get the proportion:
With your values:
Solve for x.
My solution is rather more workmanlike !
Start with the right angled triangle ABD and let the side AB = a.
Then
so
Now let the angle ACD = and use the sine rule in each of the triangles ABC and ACD.
From ACD,
so
From ABC,
so
Equate the two expessions for simplify, and you find that
and from which it follows that AD =
Call the angle ABD then from the triangle ABD, and from that you can deduce that and
Now, finally, turn your attention to the triangle at the other end, ADE.
The angle AED will equal so making use of the sine rule again,
so
so after cancelling and then substituting for and
we finish up with