Help I am totally lost. I have to do a proof problem with the statement and reason and I just don't get it.
Given: DB bisects /ADC
AD congruent CD
Prove: DB ' AC
Hi Fentonjc,
I am going to assume that /ADC means angle ADC
and DB ' AC means DB bisects AC (the only other alternative that I can think of is that DB is perpendicular to AC and the proof would be similar)
Draw the picture first of course
Let X be the intersection of DB and AC
Consider triangle ADX and triangle CDX
you can prove that these 2 triangles are congruent and hence show that AX=CX
then give a concluding statement and you are finished.
Is that enough help?
Melody.
Hello, Fentonjc!
I am getting better at Mathematical Forensics.
I think I understand the problem . . .
$\displaystyle \text{Given: }\:\text{Triangle }ADC.$
. . . . . . $\displaystyle DB\text{ bisects }\angle ADC.$
. . . . . . $\displaystyle AD = CD$
$\displaystyle \text{Prove: }\:DB \perp AC$
We are given: .$\displaystyle AD = CD.$Code:D * /|\ / | \ / | \ / | \ / | \ / | \ / | \ A * - - - * - - - * C B
We are told that $\displaystyle DB$ bisects $\displaystyle \angle ADC.$
Then: .$\displaystyle \angle BDA = \angle BDC$
Also: .$\displaystyle DB =DB$ (Identity)
So: .$\displaystyle \Delta DBA \cong \Delta DBC$ (s.a.s.)
Hence: .$\displaystyle \angle DBA = \angle DBC$ (corresponding parts)
We see: .$\displaystyle \angle DBA + \angle DBC \,=\,180^o$
Since $\displaystyle \angle DBA = \angle DBC,\;\angle DBA = 90^o$
Therefore: .$\displaystyle DB \perp AC.$
I tried to attach the actual picture. Then it has a box that has 2 sides ones says statements and the other says reasons.
it is a triangle that has a D on top and A B C on the bottom and it also has a 1 and 2 inside the triangle under the D