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Math Help - hypotenuse of a right triangle

  1. #1
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    hypotenuse of a right triangle

    Hi,

    I need help with this problem. May someone help me please?

    The Problem is: You are asked to prove the theorem given below, using the figure shown. What specifically do you need to prove in the figure?

    The median drawn to the hypotenuse of a right triangle divides it into two isosceles triangles.

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  2. #2
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    Thou art asked to show that,
    triangle BAD and triangle CAD are isoseles triangles.

    Use the following theorem.
    Theorem: Given a triangle ABC. From A draw a median to BC. Let is intersect at D.
    Then,
    AD = (1/2)*sqrt(2(AB^2+AC^2)-BC^2)

    Now the triangle which we have is right.
    Hence,
    AB^2+AC^2 = BC^2 by Pythagorean Theorem.

    Thus,
    AD=(1/2)*sqrt(2BC^2-BC^2)=(1/2)*sqrt(BC^2)=(1/2)BC
    But (1/2)BC=BD=DC
    Hence,
    AD=BD and AD=DC
    Hence the two triangles are isoseles.
    Q.E.D.
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  3. #3
    Super Member

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    Hello, JayJay!

    Prove: the median drawn to the hypotenuse of a right triangle
    divides it into two isosceles triangles.
    There is an "eyeball" proof for this.
    Maybe you can explain it in words.


    Code:
                    * * *      C
              *               o
            *             o    o*
           *          o         o*
                  o              o
          *   o                   *
        A o   o   o   *   o   o   o B
          *     r     O     r     *
    
           *                     *
            *                   *
              *               *
                    * * *

    A right triangle can be inscribed in a semicircle.

    We have right triangle ABC inscribed in a semicircle with radius r.
    . . Hence: OA = OB = r.

    Draw radius OC and we have:
    . . OC = OA . . ∆AOC is isosceles.
    . . OC = OB . . ∆COB is isosceles.

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  4. #4
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    Thank you
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