View Full Version : geometry proving (not for marks!)

red_baron

10-05-2002, 05:25 PM

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:

ABCD is a quadrilateral whose area is bisected by the diagonal AC. Prove that BD is bisected by AC.

ygfperson

10-05-2002, 05:38 PM

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.

red_baron

10-05-2002, 05:42 PM

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

Cshot

10-05-2002, 06:09 PM

>> 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.

dirkduck

10-05-2002, 06:09 PM

I dont believe thats possible with a plain quadralateral, should be "at least" a rectangle.

red_baron

10-05-2002, 06:19 PM

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

Cshot

10-05-2002, 06:31 PM

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

red_baron

10-05-2002, 06:33 PM

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:

Cshot

10-05-2002, 06:37 PM

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.

red_baron

10-05-2002, 06:44 PM

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.

MethodMan

10-05-2002, 07:31 PM

Just out of curiosty, what part of Canada are you from?

red_baron

10-05-2002, 07:34 PM

mississauga, ontario

red_baron

10-06-2002, 01:06 PM

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.

Cshot

10-06-2002, 03:09 PM

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.

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