# geometry proving (not for marks!)

• 10-05-2002
red_baron
geometry proving (not for marks!)
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.
• 10-05-2002
ygfperson
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.
• 10-05-2002
red_baron
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
• 10-05-2002
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.
• 10-05-2002
dirkduck
I dont believe thats possible with a plain quadralateral, should be "at least" a rectangle.
• 10-05-2002
red_baron
Quote:

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.

ahh i see indirect proof, thanx for the help all :D
• 10-05-2002
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

of course you'll need illustrations in your proof labelling the correct parameters
• 10-05-2002
red_baron
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:
• 10-05-2002
Cshot
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.
• 10-05-2002
red_baron
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.
• 10-05-2002
MethodMan
Just out of curiosty, what part of Canada are you from?
• 10-05-2002
red_baron
mississauga, ontario
• 10-06-2002
red_baron
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