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Brain fart - "easy" geometry problem

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Thanks for knocking a little rust off the cranial drivetrain. It is still inoperable, but it *almost* got up to speed on this one 😛
 
If the above diagrams are right, then the key is to draw an additional chord from the end if the chord to the other end of the first diagonal, making another right triangle that is similar to the one in question.

Now with solutiony goodness below
saXH1.jpg

Triangle BDE is similar to ABC, therefore,
BE/BD = BC/AB, with BC=3/4*BD
Rearranging that, you end up with the ratio of AB to BC is 2/sqrt(6). The angle is ~35 degrees per the posts below
Why is DB perpendicular to DE?
 
Because if BE is a diagonal per OP, then BDE is a semicircle and any angle whose endpoints are the ends of a diagonal, must be a right angle.
 
Because if BE is a diagonal per OP, then BDE is a semicircle and any angle whose endpoints are the ends of a diagonal, must be a right angle.

Or more generally, an inscribed angle measures one half the measure of the intercepted arc.
 
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