You are not logged in.
Pages: 1
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.
"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay
Offline
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.
i dont know................as this came in my question paper
"Let us realize that: the privilege to work is a gift, the power to work is a blessing, the love of work is success!"
- David O. McKay
Offline
△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
Offline
△ADB==△BCA then <DBA=<CAB thus PB=PA
Same prosedure PC=PDHowever, 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!"
- David O. McKay
Offline
No, sorry, SAB in a same line sequentially, SD//BC
X'(y-Xβ)=0
Offline
please explain this question answer briefly fastly.its urgent =(=(=(=(=(=(
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!"
- David O. McKay
Offline
This is the graph, follow the steps with respective color, you will find the proof.
X'(y-Xβ)=0
Offline
SBCD is a parallelogram, which has the properties that SD=BC and <S+<B=180°
X'(y-Xβ)=0
Offline
Pages: 1