Triangle split in half angle-wise
I'm searching for a simple expression for the distance CD in relation to BE (=2BD) in the picture above.
1) D is midpoint on BE. BE and hence BD = DE are known.
2) Angles BAC = CAE = BAE/2, that is, C represents the "angular midpoint" on BE as "seen from A". Those angles are known.
3) F is the center of a circle. Points A, B and E lie on the perimeter of that circle so AF = BF is its radius.
4) Angles BFD = BAE are known. Hence triangle BDF is completely known (because of right angle and BD are also known).
Number 4 above is actually necessary because the circle describes all positions of point A which yield constant angle BAE. (See PlanetMath: circumferential angle is half the corresponding central angle) I would be very interested in knowing if there exists a similar simple relationship for all points A such that instead CD remains constant!
The problem has implications for perspective. If you stand at A and look at BE, then BC will appear to be as long as CE. And I want to know what other points of observation A exist, which would remove the actual center D from the apparent center C with an equal distance CD. Alternatively, at least how CD relates to BD.
I hope that you find the question a little interesting!