Is anyone good with Geometric Proofs?
I need help!
Ill attach the paper

- anonymous

Is anyone good with Geometric Proofs?
I need help!
Ill attach the paper

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- anonymous

##### 1 Attachment

- anonymous

@Hero

- e.mccormick

First off, you know the principals relating to angles of lines that intersect?

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- anonymous

ok

- e.mccormick

Not sure if you know them or not. I mean this stuff:
|dw:1440021663729:dw|

- anonymous

I know a little, but looking at the 2nd problem why does it say PC = QB There is no P or Q

- e.mccormick

Probably a typo. I beat they meant \(AD\cong BE\).

- e.mccormick

bet...

- e.mccormick

On the first one, you can use the fact that the intersections of lines must add up to 180 degrees to get a generic measure for the angles at points B abd C. Then use the givens that the inner angles of B and C are the same to show that those outers are using 180-(the same number) to show thart they are the same. I just do not remember the names of the theroems off hand to show that, but you should have a list somewhere in their book or on ylur class web site of theorems you are using.

- e.mccormick

Something to do with a transversal making supplementary angles...

- e.mccormick

But yah, on the second one, I would just ask the teacher if it is supposed to be what it is, since those other points do not exist.

- anonymous

thank you

- e.mccormick

Ah, here, found this:
Linear Pair Postulate If two angles form a linear pair, then the measures of the angles
add up to 180°.
That looks like the one you need to use in the first one. And lets see... Triangle Exterior Angle Postulate could also do it, which says that "The measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles."
Either of those can get you in the correct direction.
I found a PDF with the different postulates and theorems.
http://web.cerritos.edu/dford/SitePages/Math_70_F13/PostulatesandTheorems.pdf

- anonymous

oh great thank you so much!!

- e.mccormick

This PDF might also help... but the visual queues did not show up =(
http://www.riverview.wednet.edu/parade/teachers/mcclintict/Geometry-1/Geo-PostulatesTheorems-List-2011.pdf

- anonymous

You are so much help

- e.mccormick

With proofs, it cames down to finding a set of rules, one after the other, that will let you say "This works!" or "This never works!" in the proper order and with the proper name. So I can tell you that when one segment touches a line, the angles on each side of where it touces add up to 180 degrees... but it does not mean much. On the other hand, if I say Linear Pair Postulate, it means exactly that but is using words that a math teacher likes. Hehe.

- anonymous

I'll have to remember that

- anonymous

what did I do wrong? :'( @e.mccormick

##### 1 Attachment

- e.mccormick

They want you to prove they are conguent segments.

- anonymous

ive tried fixing it and I can quite figure it out where I went wrong ? was it because I put dc in cd and ec instead ce

- e.mccormick

The DC = EC in step B is wrong. You want to prove that int he end. You could say AC = BC in B.

- anonymous

so b) would be ac = bc , cd = ce?

- e.mccormick

No, you are given AC cong BC, and some mistake... I am betting that becaus it says you need to prove CD cong CE that the mistake is AD cong BE. If the given is:
AC cong BC, AD cong BE then:
AC = BC, AD = BE by Def. of Cong. Segments

- anonymous

so then c is ac + ad = bc + be for addition property

- anonymous

i thought i knew what i was doing then i messed up and know im 100% lost again

- e.mccormick

OK, well, the addition property is... hmmm... not sure if you need that at all or just the segment addition postulate.
See, AD + DC = AC, which is adding up the segments to make the line. You can do the same with the other side, so BE + EC = BC.
Then you can say by substitutain that:
AD + EC = BC
For that substitution to be valid, EC = DC.
Does that make some sense?

- e.mccormick

Oh, and on the AD + EC = BC you can make it clearer by saying:
AD + EC = BC Subsitutuion of AD for BE.
Basically you are saying, "If I can subsitute the upper line segment, then I must also be able to substitute the lower line segment, therefore the lower ones are equal and equal means congruent."

- anonymous

okay let me see if i got this right
a) Ac = bc, ad=be
b)ac=bc, ad = be
c)ac +ad = bc +be
d) ad + dc =ac
be + ec = bc
e)ec =dc
f) dc = ce

- anonymous

@e.mccormick

- e.mccormick

c, as you have it, really does not need to exists.
c) ad + cd = ac, be + ec = bc, segment addition postulate
d) ad + ce = bc, subsitution of ad for be
e) ad + ce = ac, subsitution of ac for bc
Hmmm..... at that point, I know it is saying CD = CE beacuse they are in the same place int he same equation.. so probably by substitution too....

- e.mccormick

Or perhaos the converse of subsitution.

- e.mccormick

ARGH! I know what subsitution I meant!
d) ad + ce = bc, subsitution of ad for be
e) ad + ce = ac, subsitution of ac for bc
f) ad + ce = ad + cd, subsitution of ad + cd for ac
g) ce = cd... this is one of the basic addition ones. Do you remember the name of it?

- e.mccormick

Additive equality.

- anonymous

So then h) CD = ec

- e.mccormick

Yah, then you do the cd cong ce because of definition of congruent.

- e.mccormick

I hope that makes more sense now. Even I went a little sideways on that one for a moment. Well... there is usually more than one way to do it, but on this one a subsitution chain is clear. Being clear is the intent of proofs. Clear and detailed baby steps that lead to the answers is basically what a proof is.

- anonymous

Thank hopefully it's all good

- e.mccormick

And this is the full name of that additive thing...
http://www.mathwords.com/a/additive_property_of_equality.htm

- anonymous

Okay sweet

- anonymous

@e.mccormick My teacher said it was Ac=bc , Dc= ec not ad=be
im still confused

- anonymous

@e.mccormick

- anonymous

i got it!!!

- e.mccormick

Well, that would not make much differrence, and it seems you were able to apply what was done yesterday to fix it.

- anonymous

A) AC≅BC, DC≅EC
B) AC = CB, DC=CE
C) AC + CB = DC = CE
D) DC + AC = DC
CB + CE =CE
E) DC = CE
F) DC ≅ CE

- anonymous

if that isnt right then i give up on this assignment

- e.mccormick

Wait.... the teacher said it is DC≅EC at the top in the givens?

- anonymous

yes with ac = bc

- e.mccormick

Then it is stupidly simple.
A) DC≅EC
B) DC = CD, EC = CE: defenition of a line segment
C) CD≅CE: substitution of CD for DC and CE for EC.
I think your teacher made a mistake, and what we did yesterday is probably right. I do not think they would ask something that simple.
See, what I did here is the line segment from D to C is the same as from C to D based on the definition of a line segment.

- anonymous

okay so other than that part its okay?

- e.mccormick

No, I am saying that what your teacher gave you as a "correcton" means it can be done in those 3 steps.
If I go from point D to C, it is the same as if I go from C to D. Your teacher is saying that the answer is a given... which makes no sense for a proof.

- anonymous

im so confused

- anonymous

i thought my teacher wanted me to do both not just the dc=ec

- e.mccormick

Well, what we did yesterday works well... but your teacher is saying that the correction to the typo is for what you want to prove to be is as a given. That just does not make sense.

- anonymous

i dont know ill just figure it out i guess ..

- e.mccormick

What I am saying is that your teacher seems to have messed up in fixing it. It is like saying, "Prove the the road from New York to New Jersey is the same length as the road from New Jersey to New York."
Your teacher said:
Given AC≅BC, DC≅EC
Prove CD≅CE
Well duh! DC≅EC = CD≅CE
If that is NOT what your teacher meant, then the thing I went over yesterday is the correct answer.

- anonymous

Oh wow im sorry i just fully understood that
ill have to ask her about that

- e.mccormick

LOL. Yah. I think you got it yesterday and your teacher is still confused. Yesterday we basically did:
a + b = c and x + y = z
Given a = x and c = z, prove that b = y.
That is the short form, and we did all that using the geometric notation for it.

- anonymous

well ill talk to her and let know

- anonymous

let you know *

- e.mccormick

I am the dyslexic, poor typist here... so a missing word or two is no big deal. =)

- anonymous

haha she said it sounded good i just need the reasons next to those 3

- anonymous

which i got from you thank you
um do you think you can help me with the other problem on that paper?

- e.mccormick

The top one?

- anonymous

yes

- e.mccormick

The Linear Pair Postulate I pointed out yesterday, and some substitution because the inner angles are given as equal should do it.

- anonymous

my teacher said the 2 angles are given to be equal, this means it is an isosceles triangle. This should make your proof easier

- e.mccormick

Well, I do not know of any isosceles triangle specific rules that apply with how I was saying you could do it. I was just pointing out that the inner angles match and can be substituted for each other, which means you can make it so the outer ones must also subsitute... it is actually kind of similar to the second one when we did it the last way we did yesterday.

- anonymous

so A) is m

- e.mccormick

On A, notice the C and B swap positions.
A) m∠XBC = m∠XCB, Given
For B I would do:
m∠XBA = 180 - m∠XBC, and m∠XCD = 180 - m∠XCB: Linear Pair Postulate

- anonymous

what do we do for the next step im starting to get a little confused

- e.mccormick

A) m∠XBC = m∠XCB, Given
B) m∠XBA = 180 - m∠XBC, and m∠XCD = 180 - m∠XCB: Linear Pair Postulate
Two substitutions and it is done.
With what you know there, what can you subsitute in:
m∠XBA = 180 - m∠XBC

- anonymous

this problem has me completely lost

- e.mccormick

Perhaps it is all the
m∠XBA, m∠XBC etc stuff. Lets try something to make that simpler.
I will use Bi for the XBC anf Bo for XBA, that is because they are point B's inner and outer angles. Similar things can be done for point C.
|dw:1440119011210:dw|
Given: Bi = Ci
Prove: Bo = Co
A) Bi = Ci: given
B) Bo = 180 - Bi and
Co = 180 - Ci : Linear Pair Postulate
Do you see where this is going? Two substitutions and it is done.

- anonymous

yes i do understand what you did

- anonymous

@e.mccormick im just having a hard time figuring out what each step like i dont know whats next

- anonymous

i cant wrap my head around it to know what to do next but i do understand whats going on

- e.mccormick

Well, at this point it is algebra to make them look alike. You just need to do a substitution. See, the line A tells you what substitutions can be done in B.
You were given something. You use a rule to show something. Then you use the given and a new rule to substitute for the next step.

- anonymous

oh okay

- e.mccormick

Once you do that sub, then you will have one last sunstitution. The result of the first substitution and the other half of B will make for a second subsitution!
Or you could do it a little longer, just putting something on seperate lines:
Given: Bi = Ci
Prove: Bo = Co
A) Bi = Ci: given
B) Bo = 180 - Bi : Linear Pair Postulate
C) Co = 180 - Ci : Linear Pair Postulate
D) subsitution involving lines B) and A)
E) subsitution involving lines C) and D)
And it will be done! That is one operation per line, which is the normal in formal proofs.

- anonymous

okay hold im trying to figure it out lol

- anonymous

D) M

- anonymous

i think i totally messed that up @e.mccormick

- e.mccormick

The goal is to get m∠XCD = m∠XBA
So that is going to be step E no matter what. The question is, how do you get there through two substitutions.
A) m∠XBC = m∠XCB, Given
B) m∠XBA = 180 - m∠XBC: Linear Pair Postulate
C) m∠XCD = 180 - m∠XCB: Linear Pair Postulate
OK, so your next steps were:
D) m∠XBC-m∠XBC=m∠XBA=180-m∠XBC
E) ∠XBA=180-∠XBC=∠XBC-∠XCB=∠XBA=180-∠XBC
Hmm.... I am not sure how you went that way. Let me give a little more detail.
When you do a substitution, it is based on what you know. I said it involved these two lines:
A) m∠XBC = m∠XCB, Given
B) m∠XBA = 180 - m∠XBC: Linear Pair Postulate
Because of the rules of equality, only four possibilities can be true:
1) m∠XBC can be replaced by m∠XCB
2) m∠XCB can be replaced by m∠XBC
3) m∠XBA can be replaced by (180 - m∠XBC)
4) (180 - m∠XBC) can be replaced by m∠XBA
Now, while I can only replace (180 - m∠XBC) with m∠XBA, the m∠XBC INSIDE (180 - m∠XBC) can be replaced by m∠XCB.
Take a look at where that leads.

- e.mccormick

And remember, the last step is:
E) m∠XCD = m∠XBA
by some sort of substitution. So D must set up for that. It involves C and D. Well, by C:
C) m∠XCD = 180 - m∠XCB: Linear Pair Postulate
1) m∠XCD can be replaced by (180 - m∠XCB)
2) (180 - m∠XCB) can be replaced by m∠XCD
So with just a little logic, you can see that D must use at least one of those so it can become the other.

- anonymous

so D) is m∠XBC = m∠XCB, m∠XBA = 180 - m∠XBC

- anonymous

@e.mccormick

- e.mccormick

I was off... the joys of work.
C) m∠XCD = 180 - m∠XCB: Linear Pair Postulate
If you swap that m∠XCB for ... hmmm.... did I write those wrong one of these times? Argh.... Let me use the shorthand one.
Given: Bi = Ci
Prove: Bo = Co
A) Bi = Ci: given
B) Bo = 180 - Bi : Linear Pair Postulate
C) Co = 180 - Ci : Linear Pair Postulate
D) Bo = 180 - Ci : Substitute Ci for Bi.
That works because in A you say Bi is Ci.
E) Bo = Co : Substitute Co for 180 - Ci
That works becaus in C you say those are the same.
There... Now you just need to replace the shorthand with the m∠XCB stuff. I hope you can understand why it works.

- anonymous

I got D) m

- anonymous

@e.mccormick

- anonymous

D is the only im having trouble with

- e.mccormick

It would be one of the otuer ones. m

- anonymous

oh okay i think i get it now

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