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hey this is a homework question and isn't for marks (i swear!), i was wondering if anyone can help me out with it, some hints or tricks, or anything at all, help will be greatly appreaciated ;) thanx,
here it goes:
Quote:
ABCD is a quadrilateral whose area is bisected by the diagonal AC. Prove that BD is bisected by AC.
you sure it just says quadrilateral? i don't think it's provable for just quadrilaterals.
imagine a kite. two opposite corners are connected by a strip of wood. the other two corners are, also. are they equal? not by a long shot.
im positive it says quadrilateral but a specific quadrilateral who's area is split into 2 when a line is drawn from one point to the vertice on the opposite side, i just gotta prove that the other line is bisected by this line (boy i suck at explaining), i dont have to prove that the 2 triangles fromed by the diagonal are equal, thats given to me
>> you sure it just says quadrilateral? i don't think it's provable for just quadrilaterals.
Yes it's provable because of the condition that diagonal cuts the quadrilateral into two equaled area triangles.
Quite simple problem actually. Diagonal AC divides quadrilateral ABCD into two equaled area triangles ABC and ACD.
Area of both triangles = (1/2) * base * height
Both of their bases is the diagonal AC. The two heights come from the diagonal BD. If AC does not bisect BD, then their heights are not equal and thus their area would not equal either. Not really a formal proof but you get the idea.
I dont believe thats possible with a plain quadralateral, should be "at least" a rectangle.
ahh i see indirect proof, thanx for the help all :DQuote:
Originally posted by Cshot
>> you sure it just says quadrilateral? i don't think it's provable for just quadrilaterals.
Yes it's provable because of the condition that diagonal cuts the quadrilateral into two equaled area triangles.
Quite simple problem actually. Diagonal AC divides quadrilateral ABCD into two equaled area triangles ABC and ACD.
Area of both triangles = (1/2) * base * height
Both of their bases is the diagonal AC. The two heights come from the diagonal BD. If AC does not bisect BD, then their heights are not equal and thus their area would not equal either. Not really a formal proof but you get the idea.
Well it does not have to be an indirect proof.
Area of ABC = (1/2)*base*height1
Area ACD = (1/2)*base*height2
Area ABC = Area ACD
(1/2)*base*height1 = (1/2)*base*height2
height1 = height2
-> AD bisects BC
of course you'll need illustrations in your proof labelling the correct parameters
yup, thanx for da help, that was problem #12 in the geometry grade 12 book in canada, i bet the US and all aroudn the world they do it in middle school... talk about a ..........y education system here in canada :mad:
Not really. In the US they teach it starting from 9th grade on except in some special cases where they do it in middle school.
i 'guess' its not that much different casue we learned about geoemtry terms in grade 9 and did calculations now in grade 12 we are doing all the kinds of proofs.
Just out of curiosty, what part of Canada are you from?
mississauga, ontario
Quote:
Originally posted by Cshot
Well it does not have to be an indirect proof.
Area of ABC = (1/2)*base*height1
Area ACD = (1/2)*base*height2
Area ABC = Area ACD
(1/2)*base*height1 = (1/2)*base*height2
height1 = height2
-> AD bisects BC
of course you'll need illustrations in your proof labelling the correct parameters
umm i looked it over again and i noticed something if the quadrilateral is a rectangle, the hight of the two triangles are not the same as the diagonal from vertice to the opposite vertice. look at image.
Oops, I forgot about that. Been awhile.
Here's a hint: Look at those 2 new heights you've drawn with BC being the two hypothenuses of the 2 new triangles.