Question

Last, please? Thank you. :)
Two-column proof.
Given:
BA⊥BD
DF⊥BD
BC||DE
Proof:
∠1 is complementary to ∠3.

Answer 1

first,the question is incorrect,BF⊥BD is incorrect,DF⊥BD is correct,
wait a moment,

Answer 2

I will prove it :)

Answer 3

PFEEH is right !

Answer 4

wrong? Why?

Answer 5

What is the BF ?!!!

Answer 6

wait let me check

Answer 7

BF⊥BD?????think about it! It's not triangle...

Answer 8

Oh wait! You're right. It should be DF.

Answer 9

PFEH : you re right !

Answer 10

Let me change it

Answer 11

Please draw a new picture !

Answer 12

found It!!!

Answer 13

\[D _{3} = B _{2}\]
ok?

Answer 14

@stupidinmath Ok?

Answer 15

Alright.

Answer 16

No @E.ali ! Don't tell him the answer in that case!

Answer 17

No it wasnt answer !

Answer 18

Its lright. I have to do the table proof anyways

Answer 19

then,
\[90-B _{2} = 90 - D _{4}\]

Answer 20

Ok?

Answer 21

Yep, got it:)

Answer 22

so,
\[-B _{2} = -D _{4} -> B _{2} = D _{4}\]

Answer 23

we give AC || EF || BD
In ABC we have :
A=90
B+C=90
A+B+C=90
B+C = B2

Answer 24

ok?

Answer 25

@E.ali !!!!! Don't tell himthe answer!!!

Answer 26

Oh.. okay..

Answer 27

It the proof !
He should know that !
He can answer than me !
Look !>>

Answer 28

Oh Oh,sorry @stupidinmath , i made a mistake!

Answer 29

I think so, too.
And, its a "SHE"

Answer 30

Its alright:)

Answer 31

\[D _{4} = B _{1} \]

Answer 32

∠1 is complementary to ∠2
∠3 is complementary to ∠4
so we have:
∠1 is complementary to ∠3

Answer 33

OK?
got it?

Answer 34

Yes,very clearly