Geometry Proofs Help - Get Ready For Monday Exam!

[ccseagle]

Geometry Proofs.. Help!

Please please someone help me! I have a Geometry Exam on Monday and I don't understand proofs one bit ! If someone could help me with a few proofs that would be so awesome!

 

[cristo]

Please please someone help me! I have a Geometry Exam on Monday and I don't understand proofs one bit ! If someone could help me with a few proofs that would be so awesome!

Have you got any specific questions?

 

[ccseagle]

yeah, could you help me with this proof..

Given: B is between A and C; D is between C and E; and C is the midpoint of line BD and of line AE

Prove: line AB is congruent to line DE

Picture:

<--A---B---C---D---E-->

 

[ccseagle]

yeah, could you help me with this proof..

Given: B is between A and C; D is between C and E; and C is the midpoint of line BD and of line AE

Prove: line AB is congruent to line DE

Picture:

<--A---B---C---D---E-->

 

[cristo]

Ok, I'll give a hint. You know that |\vec{AC}|=|\vec{CE}| so start with this, then express both sides of the equation in terms of two new lengths. Have a go, and post your thoughts.

 

[ccseagle]

AB+BC= line AC

and

CD+CE = line CE

I don't know, am I even on the right track??

 

[cristo]

Yes, so now the equation becomes:|\vec{AC}|=|\vec{CE}| \Rightarrow<br /> |\vec{AB}|+|\vec{BC}|=|\vec{CD}|+|\vec{DE}|<br /> Carry on with this.

 

[ccseagle]

umm.. would the reason be substitution??

 

[ccseagle]

Line AC is congruent to line CE??

 

[cristo]

Ok, my notation

|\vec{AC}|=|\vec{CE}|

means that the length of the line AC is equal to the length of the line CE (this is equivalent to the two lines being congruent).

We know that AC is congruent to CE, since C is the midpoint of AE. CAn you say anything similar regarding the lines BC and CD

 

[ccseagle]

C is the midpoint of lines BC and CD?

 

[cristo]

WEll, C is the midpoint of BD, therefore BC and CD are congruent. Now look at the equation given in post #7. What does this tell you?

 

[ccseagle]

I don't think I understand.. Can you give me a hint??

 

[cristo]

I don't think I understand.. Can you give me a hint??

Not without giving you the answer! Ok, here goes: The equation can be rearranged to give

|\vec{AB}|-|\vec{DE}| =|\vec{CD}|-|\vec{BC}|
Now, since the BC and CD are congruent, what can say about the RHS of the equation? What does this then imply?

 

[ccseagle]

I don't know.. Umm what's a RHS??

 

[cristo]

Right hand side. Ok, if two lines are congruent, then it means they have the same length. Let's call this length x. Then the RHS is x-x=...?

 

[ccseagle]

the lengths are congruent?

 

[cristo]

Well, |\vec{AB}|-|\vec{DE}| =|\vec{CD}|-|\vec{BC}|=0 \Rightarrow |\vec{AB}|=|\vec{DE}|

And so, the lines AB and DE are congruent.

 

[ccseagle]

Ok, i think i get it.. Thanks.. Do you think you could help me with a couple more??

 

[cristo]

Maybe one, it's getting quite late. Post your attempt at a proof to the problem you post first though.

 

[ccseagle]

ok, thanks..

Given: angle 8 and angle 1 are supplementary

Prove: angle 4 is congruent to angle 6

Picture:


___________________________________________

-angle 1 is congruent to angle 3 because they are vertical angles
-angle 8 is congruent to angle 6 because they are vertical angles

 

[ccseagle]

-supplementary means that the sum of two angles equals 180 degrees
-angle 8 and angle 1 are congruent because of corresponding angles

 

[ccseagle]

But now is where I get stuck

 

[ccseagle]

Umm, can you give me a hint please??

 

[cristo]

-supplementary means that the sum of two angles equals 180 degrees
-angle 8 and angle 1 are congruent because of corresponding angles

Careful.. You said 8 and 1 were supplementary!

From what we are given, and your points in your previous post,

6=8=180-1

Now, since 1 and 2 are angles on a line, what can you say about 2 in terms of 1? how is 2 related to 4?

 

[ccseagle]

-angle 2 and angle 4 are vertical angles
-angle 2 and angle 1 is a linear pair

 

[ccseagle]

but also in my given it says that angle 1 and angle 8 are supplementary...

 

[cristo]

Yes, so 2=180-1 and 2=4. Put these into the eqn above

 

[ccseagle]

the equation in post #25??

and i don't understand wut u mean.. how do i put into the equation above??

 

[cristo]

yes 6=8=180-1

 

[ccseagle]

is this right?

6=8=179??

 

[cristo]

No sorry, 1 was the angle: 6=8=180-1=2=4. Hence 6 is congruent to 4.

 

[ccseagle]

ooohh.. so we solved it??

 

[cristo]

yup we sure did

 

[ccseagle]

wow! well thanks, you helped alot.. well, i guess i'll let you go now.. Thank you SO MUCH! :):):):):)

 

[cristo]

You're welcome. Hope you learn the methods to enable you to do well in your exam.

 

[gabee]

It's been so long since I took geometry...I forgot what it was like to do proofs!

 

[ccseagle]

Thanks, I think with wut you helped me with I should be able to do good.. Thanks again..

 

[ccseagle]

:).. Well they arent fun I can tell you that..

 

[cristo]

Also, for future reference, the subject of these questions is not Differential Geometry. In future, it may be better if you posted this in the Precalculus Maths section in the Homework Forum.

 

[ccseagle]

oh, :), I thought I did thanks..

 

[Dragondude]

Isoceles triangles

I'm not sure on a question. Could somone help me?

 

[Nesa55]

Ok i need a little help on this Proof.
Given: <1 and <2 are right angles
H is the midpoint of segment FK; FK ll HJ

Prove:FG is congruent to HJ
I have half of it done but I'm stuck on it. HELP!