Short Pythagorean Theorem Proof

Discussion in 'Algebra' started by nycmathguy, Sep 18, 2021.

  1. nycmathguy

    nycmathguy

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    Set 1.4

    Can you do parts (a-d)? This will be math notes for me. Thanks.

    20210917_192104.jpg
    20210917_192125.jpg
     
    nycmathguy, Sep 18, 2021
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  2. nycmathguy

    MathLover1

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    a.

    in triangle BCD and CDA

    given < CAD = < DCB
    also
    < BCA = < CDB (both 90 degrees)
    CD=CD (common side)
    =>triangles BCD and BAC are similar
    if so, then corresponding sides are proportional
    a/y=c/a

    b.

    a/y=c/a->a^2=cy


    c. Show that triangle ACD is similar to ABC , and use this to deduce that b^2=c^2-cy.

    triangle ACD is similar to ABC: all angles are congruent
    each have 90 degree angle, given < CAD = < DCB, so third angles are equal

    then sides are proportional
    b/(c-y)=c/b cross multiply
    b^2=c(c-y)
    b^2=c^2-cy

    (d)

    We need to combine the two equations in part (b) and (c) to arrive at c^2=a^2+b^2

    a^2=cy
    b^2=c^2-cy
    ............substitute cy
    b^2=c^2-a^2
    a^2+b^2=c^2
     
    MathLover1, Sep 18, 2021
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Thank you very much.
     
    nycmathguy, Sep 18, 2021
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