1. ## givens and such

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 .

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

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

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

Do you know what the "indirect method" is?

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

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

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

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