Hence
]]>Bob
]]>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?
]]>