givens and such

**tysonrss** · December 7th 2010, 08:29 AM

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

.

**Soroban** · December 7th 2010, 10:00 AM

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.

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

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

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

Thread: givens and such

LinkBack

Thread Tools

Display

givens and such

Similar Math Help Forum Discussions

A little more help with givens...

Givens. Statements, proofs

Search Tags