can someone help me with these 3 proofs, I have attached the file.. it has to get done by today! and i really appreciate it:) medals rewarded!

- anonymous

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- anonymous

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- anonymous

@roberman228

- anonymous

okay:)

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## More answers

- anonymous

yea, ive tried google

- anonymous

@whpalmer4 can you help?

- anonymous

@jim_thompson5910 can you help me with these couple of questions about geometry proofs?

- jim_thompson5910

how far did you get on #2

- anonymous

um i think i finished that one, not sure let me check

- jim_thompson5910

ok post what you got

- anonymous

im not so sure how to do it?

- jim_thompson5910

or just write it out here

- anonymous

like what do you mean?

- anonymous

# 2 im not sure what to do

- anonymous

what is the first step i need to do @jim_thompson5910 ?

- jim_thompson5910

well start with what you are given

- anonymous

so would it be
BCD=EDC
BCD=ECD

- anonymous

both are given?

- jim_thompson5910

good, start off saying those two sets of angles are congruent because that's given

- anonymous

so then it would look like this?
BCD=EDC|GIVEN
BCD=ECD|GIVEN
and then now what?

- jim_thompson5910

what segments can you say for sure are congruent

- anonymous

uuhmm not so sure.

- anonymous

i would have to say that BCD and BED are congruent right?

- jim_thompson5910

what segment is between angles BCD and EDC

- anonymous

is it just one segment?

- jim_thompson5910

yes do you see it?

- anonymous

CDE?

- jim_thompson5910

segment, not angle

- anonymous

F?

- jim_thompson5910

that's a point, not a segment

- anonymous

whats a segment?

- jim_thompson5910

there's the figure
|dw:1358891844771:dw|

- jim_thompson5910

what I'm talking about is this segment (in black)
|dw:1358891880746:dw|

- anonymous

ohhh C and D sorry lol

- jim_thompson5910

you mean segment CD right?

- anonymous

yes hhaha lol

- jim_thompson5910

that's the shared side between the two triangles that overlap

- jim_thompson5910

CD = CD by the reflexive property

- jim_thompson5910

then you can say those two overlapping triangles are congruent by the ASA property

- anonymous

okay, so what would our whole overall problem look like? ill do it, and you see if it looks correct

- jim_thompson5910

alright

- anonymous

BCD=EDC|GIVEN
BCD=ECD|GIVEN
so CD=CD by the reflexive property
the two overlapping triangles are congruent by the ASA property
would this be what it look like?

- jim_thompson5910

you'll have to be specific by actually naming the triangles

- anonymous

BCD=EDC|GIVEN
BCD=ECD|GIVEN
so CD=CD by the reflexive property
BCD and EDC are congruent by the ASA property
how about this?^^

- jim_thompson5910

you nailed it

- anonymous

awesome! okay how about the next one now:)

- jim_thompson5910

oh you might want to say triangles BCD and EDC are congruent by the ASA property
so you don't confuse them with angles

- anonymous

okay!

- jim_thompson5910

ok how do we get started on the next one

- anonymous

we have to start with what we are given.

- jim_thompson5910

exactly, so go ahead and state that

- anonymous

im not sure how that to start this one because Ive never done one with a midpoint in it, but ill try real quik

- jim_thompson5910

well you start off by stating that D is the midpoint of AC (given)
then you use the definition of midpoint to say that AD = CD

- anonymous

BDC=BDA|GIVEN
BCD=BAD|GIVEN
RIGHT?

- jim_thompson5910

where does it say that BCD=BAD is given?

- anonymous

actually no lol

- jim_thompson5910

but the first one is given

- jim_thompson5910

so add that on

- anonymous

AD=CD|GIVEN
right?

- jim_thompson5910

AD = CD not because it's given, but by the definition of midpoint

- jim_thompson5910

A midpoint cuts a line segment into two equal halves

- anonymous

so then how do i find my other given?

- jim_thompson5910

You're done with you're givens

- anonymous

oh so theirs only one given then

- jim_thompson5910

Now you need one more side or one more angle, here is a visual hint:
|dw:1358892974213:dw|

- jim_thompson5910

There are 2 givens, but you've used them up already

- anonymous

so then if we have BDC=BDA|GIVEN
I would have to look at the other angle i have on the picture to get my other given?

- jim_thompson5910

|dw:1358893179414:dw|

- anonymous

BAD=BCD ?

- jim_thompson5910

see those segments i highlighted?

- anonymous

yep

- jim_thompson5910

they are congruent, do you see why?

- anonymous

yes, because they have equal sides right?

- jim_thompson5910

they are congruent because they are the same segment BD
|dw:1358893344746:dw|

- jim_thompson5910

so you'll have to say BD = BD and give a reason why that's true

- anonymous

so I would say BD=BD are reflective property ?

- jim_thompson5910

good

- anonymous

okay and then I say why my triangles are congruent?

- jim_thompson5910

yes and which property will you use

- jim_thompson5910

this may help
|dw:1358893552844:dw|

- anonymous

so then it would look like this:
BDC=BDA|GIVEN
So BD=BD by the reflective property
Triangles BDC and BDA are congruent by the ASA property
is this right?

- jim_thompson5910

you're missing something

- jim_thompson5910

AD = DC but why is that?

- jim_thompson5910

and remember you have 2 given statements

- anonymous

because its the midpoint, where would i place it?

- jim_thompson5910

and it's not the ASA property

- anonymous

would it be the SAS property?

- jim_thompson5910

yes that's correct

- anonymous

BDC=BDA|GIVEN
AD = DC|GIVEN
So BD=BD by the reflective property
Triangles BDC and BDA are congruent by the SAS property

- jim_thompson5910

no AD = DC is not given

- jim_thompson5910

go back and look at the doc and you'll see it's not given

- anonymous

okay, so then what would our next given be and is SAS correct for the property?

- jim_thompson5910

yes it's SAS

- jim_thompson5910

look back at the doc, and simply restate what they give

- anonymous

all i see is Dis the midpoint of side AC, and then BDC=BDA

- jim_thompson5910

there you go

- jim_thompson5910

just restate those and say given

- jim_thompson5910

that will allow you say that AD = DC because of the definition of midpoint

- anonymous

so how would it look ?

- jim_thompson5910

like you had it, just add those lines in

- anonymous

okay, where would I put that at when i put my answer in?

- anonymous

so would it look like this:
BDC=BDA|GIVEN
side AD =side DC|GIVEN
So BD=BD by the reflective property
Triangles BDC and BDA are congruent by the SAS property

- jim_thompson5910

put those givens at the top, then put AD = DC | definition of midpoint right after that

- anonymous

oh okay!

- anonymous

AD = DC | definition of midpoint
BDC=BDA|GIVEN
So BD=BD by the reflective property
Triangles BDC and BDA are congruent by the SAS property

- jim_thompson5910

you forgot AD = DC | definition of midpoint again

- anonymous

BDC=BDA|GIVEN
AD=DC|GIVEN
AD = DC | definition of midpoint
So BD=BD by the reflective property
Triangles BDC and BDA are congruent by the SAS property

- jim_thompson5910

good

- anonymous

OK last one!:)

- jim_thompson5910

ok show me what you got for this one

- anonymous

kk

- anonymous

I got:
AB=CD|GIVEN
BF=ED|GIVEN
is this right?

- jim_thompson5910

AB || CD, not AB = CD

- jim_thompson5910

also, angle B = angle D | given

- anonymous

so then it would look like this
AB||CD|GIVEN
angle B=angle D|given

- jim_thompson5910

along with BF=ED|GIVEN

- jim_thompson5910

you have 3 given statements

- anonymous

oh how would i order them?

- jim_thompson5910

in any order, all 3 are given and don't depend on one another

- anonymous

AB||CD|GIVEN
angle B=angle D|given
BF=ED|GIVEN
now what?

- jim_thompson5910

because AB || CD, we can say that
angle BAF = angle DCE
because they are alternate interior angles

- jim_thompson5910

so essentially
angle BAF = angle DCE | alternate interior angles
what's next?

- jim_thompson5910

you have 2 angles, and a side, which congruence property can you use?

- anonymous

BAF=angle DCE|alternate interior angles
angle B=angle D|given
BF=ED|GIVEN
just to clear it up is this right^^ if its not can you correct it?

- jim_thompson5910

you put the givens first though
the "given" statements always come first

- jim_thompson5910

since you always start with what you are given

- jim_thompson5910

and you build from there

- anonymous

BF=ED|GIVEN
angle B=angle D|given
BAF=angle DCE|alternate interior angles

- jim_thompson5910

you forgot the line AB||CD | given

- anonymous

where do i put that that at

- jim_thompson5910

above the "alternate interior angles" line

- anonymous

BF=ED|GIVEN
angle B=angle D|given
AB||CD | given
BAF=angle DCE|alternate interior angles

- jim_thompson5910

good

- anonymous

okay now what:)

- jim_thompson5910

you have 2 angles, and a side, which congruence property can you use?

- anonymous

ASA OR AAS

- jim_thompson5910

you would use AAS because the two angles are adjacent and the last side is not between them

- anonymous

Ok, so then would i say anything about reflective property?

- jim_thompson5910

no you can't use the reflexive property for this one

- anonymous

so it looks like this:
BF=ED|GIVEN
angle B=angle D|given
AB||CD | given
BAF=angle DCE|alternate interior angles
triangle BF and ED are congruent by the AAS property

- jim_thompson5910

last line is more like
triangle ABF and triangle CDE are congruent by the AAS property

- anonymous

okay, but otherwise it looks good?

- jim_thompson5910

yes it looks great

- anonymous

thanks so much jim! you tha best:))

- jim_thompson5910

you're welcome

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