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#1 2007-02-01 03:20:09

soha
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Registered: 2006-07-07
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In A Trapezium,.....

IN A TRAPEZIUM ABCD,AB PARALLEL TO CD AND AD=BC.IF P IS THE POINT OF INTERSECTION OF THE DIAGONALS.PROVE THAT PA ×  PC = PB × PD.


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#2 2007-02-01 07:25:10

Thomas11
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Re: In A Trapezium,.....

The two condititions you made(parallel sides AB and CD, AD=BC) can have only two conclusions for the trapezium: either it's a parallelogram or it's an isosceles trapezium.
If it's an isosceles trapezium its axis-symmetric and the axis is orthogonal to AB anc CD and P is on this axis. Thus: PA=PB and PC=PD.
If it's a parallelogram, it's diagonals bisect each other at the point P. Therefore you know: 0,5AC=PA=PC and 0,5BD=PB=PD. The equation looks like this now: PA^2=PB^2 . This, however, is not to prove as this equation is false for parallelograms. The equation you made is not correct for all trapeziums you described.

#3 2007-02-01 16:06:41

soha
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Registered: 2006-07-07
Posts: 2,530

Re: In A Trapezium,.....

i dont know................as this came in my question paper


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#4 2007-02-02 14:37:33

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: In A Trapezium,.....

△ADB==△BCA then <DBA=<CAB thus PB=PA
Same prosedure PC=PD

However, to prove △ADB==△BCA, you need to prove <DAB=<CBA since we have DA=CB and AB=BA already. To prove this you need to use a parallegon SBCD just embeding ABCD and SB//CD

Last edited by George,Y (2007-02-02 14:43:08)


X'(y-Xβ)=0

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#5 2007-02-02 22:25:40

soha
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Registered: 2006-07-07
Posts: 2,530

Re: In A Trapezium,.....

George,Y wrote:

△ADB==△BCA then <DBA=<CAB thus PB=PA
Same prosedure PC=PD

However, to prove △ADB==△BCA, you need to prove <DAB=<CBA since we have DA=CB and AB=BA already. To prove this you need to use a parallegon SBCD just embeding ABCD and SB//CD

which parallelogon..and what is S in your figure..is this the point of intersection  of the diagonals which is supposed to be P ..


"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!" smile smile
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#6 2007-02-02 23:40:45

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: In A Trapezium,.....

No, sorry, SAB in a same line sequentially, SD//BC


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#7 2007-02-03 23:14:03

soha
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Registered: 2006-07-07
Posts: 2,530

Re: In A Trapezium,.....

please explain this question answer briefly fastly.its urgent sad=(=(=(=(=(=(

Last edited by soha (2007-02-03 23:14:40)


"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!" smile smile
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#8 2007-02-04 14:36:13

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: In A Trapezium,.....

This is the graph, follow the steps with respective color, you will find the proof.


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#9 2007-02-04 14:52:25

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: In A Trapezium,.....

SBCD is a parallelogram, which has the properties that SD=BC and <S+<B=180°


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