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- Agnishom
- 2013-01-21 13:47:34
Got it .... Thanks 
- scientia
- 2013-01-21 07:25:15
Hence
- bob bundy
- 2013-01-21 02:40:45
Now my diagram isn't making sense. Is the lettering A to B to C to D ?
Bob
- Agnishom
- 2013-01-21 00:25:33
Let CD < AB,
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
- bobbym
- 2013-01-21 00:18:35
Hi Agnishom;
That one I believe works, yes post the hint.
- Agnishom
- 2013-01-21 00:05:57
Sorry, it should have been AC² + BD² = BC² + AD² + 2•AB•CD
P.S.: I have a hint in my book. But I dont get it. Do you want it?
- Agnishom
- 2013-01-21 00:04:22
Let me check....
- bobbym
- 2013-01-20 23:54:30
Not so different. I blunder around plenty but Agnishom is usually careful.
- bob bundy
- 2013-01-20 22:44:41
hi bobbym
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
- bobbym
- 2013-01-20 22:37:06
Hi 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.
- bob bundy
- 2013-01-20 22:27:45
hi bobbym,
Wow! Are we in telepathic communication or what ?
My edit was at your time less 20 seconds.
Bob
- bobbym
- 2013-01-20 22:25:18
Hi Agnishom;
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?
- bob bundy
- 2013-01-20 20:53:14
hi Agnishom,
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 + 2•AB•CD
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
- Agnishom
- 2013-01-20 15:46:49
In a trapezium ABCD, if AB||CD then AC² + BD² = BC² + AB² + 2•AB•CD
How?