Proof Help asap please :)

[Jh1193]

Hey I'm stuck on a couple proofs and could really use some help.

First one goes like so:

Givenoint P is not on line ABCD and PB=PC
Prove:Angle ABP = Angle DCP

So far I have this:
Statement[reason]

1.Point P is not on line ABCD, PB=PC[given]
2.Angle PBC=Angle PCB[if 2 sides of a triangle are congruent, then the angles opposite those angles are congruent]
3.Angle ABP and Angle PBC are supp[if two angles form a linear pair, then those angles are supp.]
Angle DCP and angle PCB are supp ^same thing

Not really sure if I'm doing this right/headed in the right direction...

http://img293.imageshack.us/my.php?image=mathpictureszu9.png
^picture for that problem



Got one more...
Given:Line AEB and line CED bisect each other @ E
Prove: AC||BD
So far I have:
Statement[reason]
1.line AEB and line CED bisect each other @ E[Given]
2.AE congruent to EB[if a seg bisects another seg, then it divides the segs into 2 congruent segs]
DE congruent to EC^same thing
As far as I have gotten...not really sure how to go from there and how to prove the lines congruent...

http://img261.imageshack.us/my.php?image=mathpicture2oz0.png
^picture for that problem.


Thanks in advance!

 

[fasteddie65]

Hey I'm stuck on a couple proofs and could really use some help.

First one goes like so:

Givenoint P is not on line ABCD and PB=PC
Prove:Angle ABP = Angle DCP

So far I have this:
Statement[reason]

1.Point P is not on line ABCD, PB=PC[given]
2.Angle PBC=Angle PCB[if 2 sides of a triangle are congruent, then the angles opposite those angles are congruent]
3.Angle ABP and Angle PBC are supp[if two angles form a linear pair, then those angles are supp.]
Angle DCP and angle PCB are supp ^same thing

Not really sure if I'm doing this right/headed in the right direction...

http://img293.imageshack.us/my.php?image=mathpictureszu9.png
^picture for that problem"

Usually, when a negative statement is to be proven, you should use an indirect proof, i.e. assume the opposite of that hypothesis.



Got one more...
Given:Line AEB and line CED bisect each other @ E
Prove: AC||BD
So far I have:
Statement[reason]
1.line AEB and line CED bisect each other @ E[Given]
2.AE congruent to EB[if a seg bisects another seg, then it divides the segs into 2 congruent segs]
DE congruent to EC^same thing
As far as I have gotten...not really sure how to go from there and how to prove the lines congruent...

http://img261.imageshack.us/my.php?image=mathpicture2oz0.png
^picture for that problem.

Try to use congruent triangles and then CPCTC (corresponding parts of congruent triangles are congruent).
Thanks in advance!

 

[Deleted member 4993]

Hey I'm stuck on a couple proofs and could really use some help.

First one goes like so:

Givenoint P is not on line ABCD and PB=PC
Prove:Angle ABP = Angle DCP

So far I have this:
Statement[reason]

1.Point P is not on line ABCD, PB=PC[given]
2.Angle PBC=Angle PCB[if 2 sides of a triangle are congruent, then the angles opposite those angles are congruent]
3.Angle ABP and Angle PBC are supp[if two angles form a linear pair, then those angles are supp.]

You are going in the right direction

4. PBA + PBC = 180° = PCB + DCP

and continue...



Not really sure if I'm doing this right/headed in the right direction...

http://img293.imageshack.us/my.php?image=mathpictureszu9.png
^picture for that problem



Got one more...
Given:Line AEB and line CED bisect each other @ E
Prove: AC||BD
So far I have:
Statement[reason]
1.line AEB and line CED bisect each other @ E[Given]
2.AE congruent to EB[if a seg bisects another seg, then it divides the segs into 2 congruent segs]
DE congruent to EC^same thing

3. angle AEC = angle DEB --- vertical angles

Now what can you say about triangles AEC and DEB? Does SAS ring a bell?

http://img261.imageshack.us/my.php?image=mathpicture2oz0.png
^picture for that problem.


Thanks in advance!