• December 2nd 2013, 04:58 AM
Fentonjc
Help I am totally lost. I have to do a proof problem with the statement and reason and I just don't get it.

Prove: DB ' AC
• December 2nd 2013, 05:26 AM
Melody2
Quote:

Originally Posted by Fentonjc
Help I am totally lost. I have to do a proof problem with the statement and reason and I just don't get it.

Prove: DB ' AC

I am going to demonstrate my lack of knowledge immediately by asking you to explain what the symbols mean. Please.
That is /ADC what is the forward slash
and
what is the apostrophy.
• December 2nd 2013, 05:55 AM
Melody2
Quote:

Originally Posted by Fentonjc

Prove: DB ' AC

Hi Fentonjc,

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.
• December 2nd 2013, 08:00 AM
Soroban
Hello, Fentonjc!

I am getting better at Mathematical Forensics.
I think I understand the problem . . .

Quote:

$\text{Given: }\:\text{Triangle }ADC.$
. . . . . . $DB\text{ bisects }\angle ADC.$
. . . . . . $AD = CD$
$\text{Prove: }\:DB \perp AC$

Code:

              D               *             /|\             / | \           /  |  \           /  |  \         /    |    \         /    |    \       /      |      \     A * - - - * - - - * C               B
We are given: . $AD = CD.$

We are told that $DB$ bisects $\angle ADC.$
Then: . $\angle BDA = \angle BDC$

Also: . $DB =DB$ (Identity)

So: . $\Delta DBA \cong \Delta DBC$ (s.a.s.)

Hence: . $\angle DBA = \angle DBC$ (corresponding parts)

We see: . $\angle DBA + \angle DBC \,=\,180^o$

Since $\angle DBA = \angle DBC,\;\angle DBA = 90^o$

Therefore: . $DB \perp AC.$
• December 2nd 2013, 06:09 PM
Fentonjc