Shortest Proofs of the Pythagorean Theorem

Discussion in 'Other Pre-University Math' started by nycmathguy, Aug 11, 2021.

  1. nycmathguy

    nycmathguy

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    Set 1.4
    David Cohen
    Question 32

    See figure below first attachment. I would like all four parts solved as additional math notes for self-study.

    20210810_195307.jpg

    20210810_195350.jpg
     
    nycmathguy, Aug 11, 2021
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  2. nycmathguy

    MathLover1

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    a)
    the following is sufficient to say that two right triangles are similar:
    If all three angles of a triangle are congruent but the sides are not, then one of the triangles is a scaled up version of the other. When this happens the proportions between the sides still remains unchanged which is the criteria for similarity.

    if <, CAD = <DCB, and <ADC=<BDC=90=> then <ACD=<DCB; so, all three angles of a triangle are congruent =>two right triangles BCD and BAC are similar and corresponding sides are proportional

    b)
    corresponding sides in triangles BCD and BAC are:
    a and c (hypothenuses)
    y and a (legs)

    then
    a/y=c/a ...........cross multiply to solve for a
    a*a=c*y
    a^2=cy

    c) show that triangles ACD and ABC are similar
    <ADC=<BCA=90
    <, CAD = <DCB=>then <ACD=<DCB; so, all three angles of a triangle are congruent =>two right triangles ACD and ABC are similar and corresponding sides are proportional

    corresponding sides in triangles ACD and ABC are:
    b and c (hypothenuses)
    c-y and b (legs)

    b/c=(c-y)/b
    b*b=c(c-y)
    b^2=c^2-cy

    d)
    from part b) we have a^2=cy.....eq.1
    from part c) we have b^2=c^2-cy........eq.2

    b^2=c^2-cy......eq.2..........substitute cy from eq.1

    b^2=c^2-a^2.............move a^2 to the left
    a^2+ b^2=c^2
     
    MathLover1, Aug 11, 2021
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  3. nycmathguy

    nycmathguy

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    Wow! I'm impressed. You are a true mathematician.
     
    nycmathguy, Aug 11, 2021
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  4. nycmathguy

    nycmathguy

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    Can you please check out my thread Triangle ABC?
    Your help is amazing. I thank you for taking this journey with me.
     
    nycmathguy, Aug 12, 2021
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