Hello, greatwhiteshark!
In an Indirect Proof, we work 'backwards'.
We assume the <u>opposite</u> of the punchline
. . and hope to get a contradiction of the given.
Given: Triangle ABC is scalene.
Segment BD bisects angle ABC.
Prove: Segment BD is NOT perpendicular to segment AC.
By definition, a scalene triangle has three <u>unequal</u> sides and angles.
We
assume that BD
<u>is</u> perpendicular to AC.
1. <u>/</u>BDA = <u>/</u>BDC = 90<sup>o</sup>
. . . . . . . 1. definition of perpendicular
2. <u>/</u>ABD = <u>/</u>CBD
. . . . . . . . . . . . .2. Given (BD bisects <u>/</u>ABC)
3. BD = BD
. . . . . . . . . . . . . . . 3. Identity
4. Δ BDA congruent Δ BDC
. . . 4. a.s.a.
5. AB = BC
. . . . . . . . . . . . . . . 5. 'corresponding parts'
But this means that triangle ABC is <u>isosceles</u> (two equal sides),
. . and we know that the triangle is scalene (<u>no</u> equal sides).
We have reached a contradiction.
. . Hence, our assumption was incorrect.
Therefore, BD is
not perpendicular to AC.