anonymous
  • anonymous
Last, please? Thank you. :) Two-column proof. Given: BA⊥BD DF⊥BD BC||DE Proof: ∠1 is complementary to ∠3.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1377959905491:dw|
anonymous
  • anonymous
first,the question is incorrect,BF⊥BD is incorrect,DF⊥BD is correct, wait a moment,
anonymous
  • anonymous
I will prove it :)

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More answers

anonymous
  • anonymous
PFEEH is right !
anonymous
  • anonymous
wrong? Why?
anonymous
  • anonymous
What is the BF ?!!!
anonymous
  • anonymous
wait let me check
anonymous
  • anonymous
BF⊥BD?????think about it! It's not triangle...
anonymous
  • anonymous
Oh wait! You're right. It should be DF.
anonymous
  • anonymous
PFEH : you re right !
anonymous
  • anonymous
Let me change it
anonymous
  • anonymous
Please draw a new picture !
anonymous
  • anonymous
anonymous
  • anonymous
found It!!!
anonymous
  • anonymous
\[D _{3} = B _{2}\] ok?
anonymous
  • anonymous
@stupidinmath Ok?
anonymous
  • anonymous
|dw:1062298227889:dw|
anonymous
  • anonymous
Alright.
anonymous
  • anonymous
No @E.ali ! Don't tell him the answer in that case!
anonymous
  • anonymous
No it wasnt answer !
anonymous
  • anonymous
Its lright. I have to do the table proof anyways
anonymous
  • anonymous
then, \[90-B _{2} = 90 - D _{4}\]
anonymous
  • anonymous
Ok?
anonymous
  • anonymous
Yep, got it:)
anonymous
  • anonymous
so, \[-B _{2} = -D _{4} -> B _{2} = D _{4}\]
anonymous
  • anonymous
we give AC || EF || BD In ABC we have : A=90 B+C=90 A+B+C=90 B+C = B2
anonymous
  • anonymous
ok?
anonymous
  • anonymous
@E.ali !!!!! Don't tell himthe answer!!!
anonymous
  • anonymous
Oh.. okay..
anonymous
  • anonymous
It the proof ! He should know that ! He can answer than me ! Look !>>
anonymous
  • anonymous
Oh Oh,sorry @stupidinmath , i made a mistake!
anonymous
  • anonymous
I think so, too. And, its a "SHE"
anonymous
  • anonymous
Its alright:)
anonymous
  • anonymous
\[D _{4} = B _{1} \]
anonymous
  • anonymous
∠1 is complementary to ∠2 ∠3 is complementary to ∠4 so we have: ∠1 is complementary to ∠3
anonymous
  • anonymous
OK? got it?
anonymous
  • anonymous
Yes,very clearly
anonymous
  • anonymous
@:PFEH.1999:What do you saying ?
anonymous
  • anonymous
but idk the reasons of the proof of the table :|
anonymous
  • anonymous
what?
anonymous
  • anonymous
I proved it!
anonymous
  • anonymous
do you want me to write all of that in one commant?
anonymous
  • anonymous
@stupidinmath: Always we have that I sayed >.
anonymous
  • anonymous
@PFEH.1999 no, no, its alright. Thank you for the help:)
anonymous
  • anonymous
your welcome my dear :)
anonymous
  • anonymous
@E.ali what do you mean?
anonymous
  • anonymous
Look :
anonymous
  • anonymous
|dw:1062299302748:dw|
anonymous
  • anonymous
If we have this always : a+b = c2 I think you answer your quetion fall again !
anonymous
  • anonymous
I answered it wrong?
anonymous
  • anonymous
Hey ! Can you answer with my pictures?
anonymous
  • anonymous
|dw:1377962519490:dw| This?
anonymous
  • anonymous
Yeah !

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