geometry proving (not for marks!)

This is a discussion on geometry proving (not for marks!) within the A Brief History of Cprogramming.com forums, part of the Community Boards category; hey this is a homework question and isn't for marks (i swear!), i was wondering if anyone can help me ...

  1. #1
    Registered User red_baron's Avatar
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    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:
    ABCD is a quadrilateral whose area is bisected by the diagonal AC. Prove that BD is bisected by AC.
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  2. #2
    Just because ygfperson's Avatar
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    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.

  3. #3
    Registered User red_baron's Avatar
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    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
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  4. #4
    Green Member Cshot's Avatar
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    >> 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.
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  5. #5
    Registered User dirkduck's Avatar
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    I dont believe thats possible with a plain quadralateral, should be "at least" a rectangle.

  6. #6
    Registered User red_baron's Avatar
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    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
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  7. #7
    Green Member Cshot's Avatar
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    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
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  8. #8
    Registered User red_baron's Avatar
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    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
    ¿Red Baron?

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  9. #9
    Green Member Cshot's Avatar
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    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.
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  10. #10
    Registered User red_baron's Avatar
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    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.
    ¿Red Baron?

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  11. #11
    Microsoft. Who? MethodMan's Avatar
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    Just out of curiosty, what part of Canada are you from?
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  12. #12
    Registered User red_baron's Avatar
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    mississauga, ontario
    ¿Red Baron?

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  13. #13
    Registered User red_baron's Avatar
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    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.
    Attached Images Attached Images  
    ¿Red Baron?

    "Imagination is more important than knowledge"
    -Albert Einstein (1879-1955)

    Check out my games!

    [code] /* dont forget code tags! */ [/code]

  14. #14
    Green Member Cshot's Avatar
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    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.
    Try not.
    Do or do not.
    There is no try.

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