anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
@roberman228
anonymous
  • anonymous
okay:)

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anonymous
  • anonymous
yea, ive tried google
anonymous
  • anonymous
@whpalmer4 can you help?
anonymous
  • anonymous
@jim_thompson5910 can you help me with these couple of questions about geometry proofs?
jim_thompson5910
  • jim_thompson5910
how far did you get on #2
anonymous
  • anonymous
um i think i finished that one, not sure let me check
jim_thompson5910
  • jim_thompson5910
ok post what you got
anonymous
  • anonymous
im not so sure how to do it?
jim_thompson5910
  • jim_thompson5910
or just write it out here
anonymous
  • anonymous
like what do you mean?
anonymous
  • anonymous
# 2 im not sure what to do
anonymous
  • anonymous
what is the first step i need to do @jim_thompson5910 ?
jim_thompson5910
  • jim_thompson5910
well start with what you are given
anonymous
  • anonymous
so would it be BCD=EDC BCD=ECD
anonymous
  • anonymous
both are given?
jim_thompson5910
  • jim_thompson5910
good, start off saying those two sets of angles are congruent because that's given
anonymous
  • anonymous
so then it would look like this? BCD=EDC|GIVEN BCD=ECD|GIVEN and then now what?
jim_thompson5910
  • jim_thompson5910
what segments can you say for sure are congruent
anonymous
  • anonymous
uuhmm not so sure.
anonymous
  • anonymous
i would have to say that BCD and BED are congruent right?
jim_thompson5910
  • jim_thompson5910
what segment is between angles BCD and EDC
anonymous
  • anonymous
is it just one segment?
jim_thompson5910
  • jim_thompson5910
yes do you see it?
anonymous
  • anonymous
CDE?
jim_thompson5910
  • jim_thompson5910
segment, not angle
anonymous
  • anonymous
F?
jim_thompson5910
  • jim_thompson5910
that's a point, not a segment
anonymous
  • anonymous
whats a segment?
jim_thompson5910
  • jim_thompson5910
there's the figure |dw:1358891844771:dw|
jim_thompson5910
  • jim_thompson5910
what I'm talking about is this segment (in black) |dw:1358891880746:dw|
anonymous
  • anonymous
ohhh C and D sorry lol
jim_thompson5910
  • jim_thompson5910
you mean segment CD right?
anonymous
  • anonymous
yes hhaha lol
jim_thompson5910
  • jim_thompson5910
that's the shared side between the two triangles that overlap
jim_thompson5910
  • jim_thompson5910
CD = CD by the reflexive property
jim_thompson5910
  • jim_thompson5910
then you can say those two overlapping triangles are congruent by the ASA property
anonymous
  • anonymous
okay, so what would our whole overall problem look like? ill do it, and you see if it looks correct
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • 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
  • jim_thompson5910
you'll have to be specific by actually naming the triangles
anonymous
  • 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
  • jim_thompson5910
you nailed it
anonymous
  • anonymous
awesome! okay how about the next one now:)
jim_thompson5910
  • 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
  • anonymous
okay!
jim_thompson5910
  • jim_thompson5910
ok how do we get started on the next one
anonymous
  • anonymous
we have to start with what we are given.
jim_thompson5910
  • jim_thompson5910
exactly, so go ahead and state that
anonymous
  • 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
  • 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
  • anonymous
BDC=BDA|GIVEN BCD=BAD|GIVEN RIGHT?
jim_thompson5910
  • jim_thompson5910
where does it say that BCD=BAD is given?
anonymous
  • anonymous
actually no lol
jim_thompson5910
  • jim_thompson5910
but the first one is given
jim_thompson5910
  • jim_thompson5910
so add that on
anonymous
  • anonymous
AD=CD|GIVEN right?
jim_thompson5910
  • jim_thompson5910
AD = CD not because it's given, but by the definition of midpoint
jim_thompson5910
  • jim_thompson5910
A midpoint cuts a line segment into two equal halves
anonymous
  • anonymous
so then how do i find my other given?
jim_thompson5910
  • jim_thompson5910
You're done with you're givens
anonymous
  • anonymous
oh so theirs only one given then
jim_thompson5910
  • jim_thompson5910
Now you need one more side or one more angle, here is a visual hint: |dw:1358892974213:dw|
jim_thompson5910
  • jim_thompson5910
There are 2 givens, but you've used them up already
anonymous
  • 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
  • jim_thompson5910
|dw:1358893179414:dw|
anonymous
  • anonymous
BAD=BCD ?
jim_thompson5910
  • jim_thompson5910
see those segments i highlighted?
anonymous
  • anonymous
yep
jim_thompson5910
  • jim_thompson5910
they are congruent, do you see why?
anonymous
  • anonymous
yes, because they have equal sides right?
jim_thompson5910
  • jim_thompson5910
they are congruent because they are the same segment BD |dw:1358893344746:dw|
jim_thompson5910
  • jim_thompson5910
so you'll have to say BD = BD and give a reason why that's true
anonymous
  • anonymous
so I would say BD=BD are reflective property ?
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
okay and then I say why my triangles are congruent?
jim_thompson5910
  • jim_thompson5910
yes and which property will you use
jim_thompson5910
  • jim_thompson5910
this may help |dw:1358893552844:dw|
anonymous
  • 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
  • jim_thompson5910
you're missing something
jim_thompson5910
  • jim_thompson5910
AD = DC but why is that?
jim_thompson5910
  • jim_thompson5910
and remember you have 2 given statements
anonymous
  • anonymous
because its the midpoint, where would i place it?
jim_thompson5910
  • jim_thompson5910
and it's not the ASA property
anonymous
  • anonymous
would it be the SAS property?
jim_thompson5910
  • jim_thompson5910
yes that's correct
anonymous
  • 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
  • jim_thompson5910
no AD = DC is not given
jim_thompson5910
  • jim_thompson5910
go back and look at the doc and you'll see it's not given
anonymous
  • anonymous
okay, so then what would our next given be and is SAS correct for the property?
jim_thompson5910
  • jim_thompson5910
yes it's SAS
jim_thompson5910
  • jim_thompson5910
look back at the doc, and simply restate what they give
anonymous
  • anonymous
all i see is Dis the midpoint of side AC, and then BDC=BDA
jim_thompson5910
  • jim_thompson5910
there you go
jim_thompson5910
  • jim_thompson5910
just restate those and say given
jim_thompson5910
  • jim_thompson5910
that will allow you say that AD = DC because of the definition of midpoint
anonymous
  • anonymous
so how would it look ?
jim_thompson5910
  • jim_thompson5910
like you had it, just add those lines in
anonymous
  • anonymous
okay, where would I put that at when i put my answer in?
anonymous
  • 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
  • jim_thompson5910
put those givens at the top, then put AD = DC | definition of midpoint right after that
anonymous
  • anonymous
oh okay!
anonymous
  • 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
  • jim_thompson5910
you forgot AD = DC | definition of midpoint again
anonymous
  • 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
  • jim_thompson5910
good
anonymous
  • anonymous
OK last one!:)
jim_thompson5910
  • jim_thompson5910
ok show me what you got for this one
anonymous
  • anonymous
kk
anonymous
  • anonymous
I got: AB=CD|GIVEN BF=ED|GIVEN is this right?
jim_thompson5910
  • jim_thompson5910
AB || CD, not AB = CD
jim_thompson5910
  • jim_thompson5910
also, angle B = angle D | given
anonymous
  • anonymous
so then it would look like this AB||CD|GIVEN angle B=angle D|given
jim_thompson5910
  • jim_thompson5910
along with BF=ED|GIVEN
jim_thompson5910
  • jim_thompson5910
you have 3 given statements
anonymous
  • anonymous
oh how would i order them?
jim_thompson5910
  • jim_thompson5910
in any order, all 3 are given and don't depend on one another
anonymous
  • anonymous
AB||CD|GIVEN angle B=angle D|given BF=ED|GIVEN now what?
jim_thompson5910
  • jim_thompson5910
because AB || CD, we can say that angle BAF = angle DCE because they are alternate interior angles
jim_thompson5910
  • jim_thompson5910
so essentially angle BAF = angle DCE | alternate interior angles what's next?
jim_thompson5910
  • jim_thompson5910
you have 2 angles, and a side, which congruence property can you use?
anonymous
  • 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
  • jim_thompson5910
you put the givens first though the "given" statements always come first
jim_thompson5910
  • jim_thompson5910
since you always start with what you are given
jim_thompson5910
  • jim_thompson5910
and you build from there
anonymous
  • anonymous
BF=ED|GIVEN angle B=angle D|given BAF=angle DCE|alternate interior angles
jim_thompson5910
  • jim_thompson5910
you forgot the line AB||CD | given
anonymous
  • anonymous
where do i put that that at
jim_thompson5910
  • jim_thompson5910
above the "alternate interior angles" line
anonymous
  • anonymous
BF=ED|GIVEN angle B=angle D|given AB||CD | given BAF=angle DCE|alternate interior angles
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
okay now what:)
jim_thompson5910
  • jim_thompson5910
you have 2 angles, and a side, which congruence property can you use?
anonymous
  • anonymous
ASA OR AAS
jim_thompson5910
  • jim_thompson5910
you would use AAS because the two angles are adjacent and the last side is not between them
anonymous
  • anonymous
Ok, so then would i say anything about reflective property?
jim_thompson5910
  • jim_thompson5910
no you can't use the reflexive property for this one
anonymous
  • 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
  • jim_thompson5910
last line is more like triangle ABF and triangle CDE are congruent by the AAS property
anonymous
  • anonymous
okay, but otherwise it looks good?
jim_thompson5910
  • jim_thompson5910
yes it looks great
anonymous
  • anonymous
thanks so much jim! you tha best:))
jim_thompson5910
  • jim_thompson5910
you're welcome

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