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Math Help - givens and such

  1. #1
    Junior Member
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    givens and such

    givens and such-sub9-11.jpg

    Given: BC is not congruent to BA; BD⊥CA
    Prove: DC is not congruent to DA
    Prove by the indirect method

    Another thing im no good at .
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  2. #2
    Super Member

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    Hello, tysonrss!

    \text{Given: }BC \ne BA,\;\;BD \perp CA

    \text{Prove: }\:DC \ne DA

    \text{Prove by the indirect method}
    Code:
                      B
                      *
                    * |*
                  *   | *
                *     |  *
              *       |   *
            *         |    *
        C * - - - - - * - - * A
                      D

    Do you know what the "indirect method" is?

    Assume DC \,=\,DA . . . and hope for a contradiction.


    \begin{array}{cccccccc}<br /> <br />
1. & DC \,=\,DA && 1. & \text{Given} \\ \\[-3mm]<br />
2. & BD \perp CA && 2. & \text{Given} \\ \\[-3mm]<br />
3. & \angle BDC \,=\,\angle BDA && 3. &\text{Both right angles.} \\ \\[-3mm]<br />
4. & BD \,=\, BD && 4. & \text{Identity postulate} \\ \\[-3mm]<br />
5. & \Delta BDC \cong \Delta BDA && 5. & \text{s. a. s.} \\ \\[-3mm]<br />
6. & BC \,=\,BA && 6. & \text{Corres. parts} \end{array}


    Since BC \,\ne\,BA, we have reached a contradiction.

    . . Therefore: . DC \:\ne \:DA

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