Hence

]]>Bob

]]>Drop perpendiculars DE and CF from D and C respectively.

In ΔABC, angle B is acute

Therefore, AC^2 = BC^2 + AB^2 - 2*AB*AE

In ΔABD, angle A is acute

Therefore, BD^2 = AD^2 + AB^2 - 2*AB*AF

That one I believe works, yes post the hint.

]]>P.S.: I have a hint in my book. But I dont get it. Do you want it?

]]>I thought maybe I was doing something wrong.

That's the difference between us then. I tend to blunder in and post anyway. I guess that's why my posts are full of corrections.

Four applications of the cosine rule does allow you to eliminate any angles and get an equation. It isn't that one; not even close.

That's as far as I've got.

Bob

]]>Looks that way!

I have been looking at it for a long time, I wanted to be sure. I thought maybe I was doing something wrong.

]]>Wow! Are we in telepathic communication or what ?

My edit was at your time less 20 seconds.

Bob

]]>How?

You might have a hard time proving that because I have been playing with trapeziums in geogebra and I can not find a single one where that is true.

Did you copy the problem correctly? Or what am I missing?

]]>I haven't fully explored this yet; maybe later; but it has the look of several applications of the cosine rule. As no cos(X) terms exist they must be cancelling out. Remember that angle ABD = angle BDC because of the parallels.

http://www.mathsisfun.com/algebra/trig-cosine-law.html

So perhaps you could have a go yourself ?

LATER EDIT:

I've tried what I suggested and reached a formula connecting AC, BD, BC, AB, CD. But it isn't this one:

AC² + BD² = BC² + AB² + 2ABCD

So I set up the diagram on Sketchpad and got (for my trapezium)

LHS = 162.86 and RHS = 193.64

Are you sure you've got the right identity ?

Bob

]]>How?

]]>