How do I find the altitude of a orthodiagonal right trapezoid?

well i saw this written by another guy, its sum of two bases over 2..

maybe use similar triangles to prove? IDK!

Before reading this answer, you might want to take a look at 

https://web2.0calc.com/questions/geometry-trapezoid

 

An orthodiagonal right trapezoid essentially means a trapezoid in where the diagonals intersect at a right angle.

 

Will do an answer when I can get a camera to work.

I was unable to get a camera to work, so You'll have to read carfully.

 

1.Construct a trapezoid with the description of the problem.

2.Name the verticies starting from the top left side, going clockwise, A,B,C and D respectively.

3.Draw altitudes going down from A and B. Name the points where they intersect DC F and E, F is on the left.

4.Let side \(\overline{AB}\) be a, and \(\overline{DC}\) be b. \(\overline{FE}\) is a, and \(\overline{EC}\) = \(\overline{DF}\) = \(\frac{b-a}{2}\).

5.Let the intersection of diagonals be O.

6.\(\triangle DOC\) is a 45-45-90 triangle.

 

See if you can work it out from there.