prove that in an isosceles trapezium

[sum of parallel sides] x [sum of non parallel sides]= product of diagnols

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- Nov 6th 2009, 09:21 PMjashansinghalisoscels trapezium
prove that in an isosceles trapezium

[sum of parallel sides] x [sum of non parallel sides]= product of diagnols - Nov 7th 2009, 02:09 AMBacteriusReply
*A isosceles trapezium named ABCD has just been created from nothing.*The sides (AB) and (CD) are parallel. Therefore, we have :

$\displaystyle (AB + CD)(BC + AD) = AC \times BD$

Note that BC = AD (since the trapezium is isosceles)

Therefore :

$\displaystyle (AB + CD)(2BC) = AC \times BD$

Now draw such a trapezium and think of as many useful things as you can : Thales, Pythagoras, Trigonometry, Vectors, anything. Then try putting it all together to substitute formulas into the original equation, so as to prove that the sum of the parallel sides times the sum of the non-parallel sides equals the product of the diagonals. Does it help ? - Nov 7th 2009, 08:24 PMjashansinghal
sorry,I am not able to understand

- Nov 7th 2009, 08:37 PMBacterius
What don't you understand ?

- Nov 7th 2009, 08:48 PMjashansinghal
what do you mean to say?

- Nov 7th 2009, 08:53 PMBacterius
- Nov 7th 2009, 09:02 PMjashansinghal
please help me

- Nov 7th 2009, 09:05 PMWilmer
- Nov 16th 2009, 07:29 PMjashansinghal
ya...it should be this...but how to proceed

- Nov 16th 2009, 08:11 PMBacterius
- Nov 17th 2009, 06:49 AMjashansinghal
prove that in an isosceles trapezium

[sum of parallel sides] x [sum of non parallel sides]= product of diagnols - Nov 17th 2009, 03:12 PMBacterius
- Nov 18th 2009, 07:54 AMjashansinghal
please help me....can you do first few steps

- Nov 21st 2009, 02:16 PMaidan