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## Topic review (newest first)

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?