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Math Help - An analytical geometry question!

  1. #1
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    An analytical geometry question!

    Hey guys! So I have this question in my worksheet and for some reason it sounds completely foreign to me. I think my teacher overlooked this kind of question, so I'm pretty stuck...

    "Given right triangle ABC with D the midpoint of BA.. Show that D is equidistant from B and C."

    and here is a "lovely" illustration to go with it!

    So yeah, any help would be greatly appreciated. Thanks so much!
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  2. #2
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    Quote Originally Posted by Shnub View Post
    Hey guys! So I have this question in my worksheet and for some reason it sounds completely foreign to me. I think my teacher overlooked this kind of question, so I'm pretty stuck...

    "Given right triangle ABC with D the midpoint of BA.. Show that D is equidistant from B and C."

    and here is a "lovely" illustration to go with it!

    So yeah, any help would be greatly appreciated. Thanks so much!

    Hello , Shnub

    We first confirm the coordinates of point  D , by using mid-pt formula we have  D = D\left( \frac{a+0}{2} , \frac{0+b}{2} \right) = D\left( \frac{a}{2} , \frac{b}{2} \right)

    Then , use the distance formula , the distance from D to the right angle C is :

     \sqrt{ (\frac{a}{2} - 0)^2 + ( \frac{b}{2} - 0)^2 } = \sqrt{( \frac{a}{2})^2 + (\frac{b}{2})^2 }


    and the distance  AD is  \sqrt{ (a - \frac{a}{2} )^2 + (0 - \frac{b}{2})^2 } = \sqrt{ ( \frac{a}{2})^2 + (\frac{b}{2})^2 }

    Can you see something special ?
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  3. #3
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    Hello, Shnub!

    Given right triangle ABC with D the midpoint of hypotenuse AB,
    show that D is equidistant from B and C.
    Code:
                  * * *
              *           *   C
            *               o
           *              *  *
                        *
          *           *       *
        A o - - - - o - - - - o B
          *         D         *
    
           *                 *
            *               *
              *           *
                  * * *

    Draw AC and BC.

    A right triangle can be inscribed in a semicircle.

    The midpoint of the hypotenuse is the center of the circle.

    Therefore: . DA = DB = DC = \text{radius}

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