Proportionality Theorem
I have the first column of the proof I just need help with the second.

- anonymous

Proportionality Theorem
I have the first column of the proof I just need help with the second.

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- schrodinger

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- anonymous

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- anonymous

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- anonymous

@jim_thompson5910

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## More answers

- anonymous

the first two are most likely incorrect

- anonymous

Any ideas? :/

- jim_thompson5910

why do you think the first two are incorrect?

- anonymous

I am over all unsure about proofs. I am horrible when it comes to them

- anonymous

oh wow.

- anonymous

guess I was right :p

- jim_thompson5910

right in that the first two lines of the 2 column proof are incorrect? Or that they are correct?

- anonymous

They are correct. :p

- anonymous

2 down 4 more lines to figure out .-.

- jim_thompson5910

why is angle ABC congruent to angle ADE ?

- anonymous

I dont know. because they're congruent angles?

- jim_thompson5910

That's circular reasoning and flawed. You cannot make a statement and then back up that statement with the same statement.

- jim_thompson5910

see this page
https://www.mathsisfun.com/geometry/parallel-lines.html

- anonymous

ugh. I hate proofs. Does it have to do with Bisectors?

- jim_thompson5910

go ahead and look through the link I posted
be sure to play with the interactive applet

- anonymous

They're corresponding angles?

- jim_thompson5910

very good

- jim_thompson5910

|dw:1433203991985:dw|

- jim_thompson5910

|dw:1433204002334:dw|

- jim_thompson5910

what about the next line?

- anonymous

one sec lemme fill in the third

- anonymous

Alrighty. give me a second to look on that link

- anonymous

this one confuses me.. possibly consecutive interior
angles?

- jim_thompson5910

why?

- anonymous

theres angles in and outside.. kind of.

- jim_thompson5910

|dw:1433204262730:dw|

- jim_thompson5910

we are given that DE || BC
|dw:1433204337699:dw|

- jim_thompson5910

and as you pointed out
\[\Large \sphericalangle ABC \cong \sphericalangle ADE\]
because they are corresponding angles
|dw:1433204439413:dw|

- anonymous

Alright...

- jim_thompson5910

The same applies for angle ACB and angle AED
\[\Large \sphericalangle ACB \cong \sphericalangle AED\]
because they are corresponding angles
|dw:1433204518237:dw|

- anonymous

soo corresponding angles again?

- jim_thompson5910

correct

- anonymous

Now the 5th line to me makes no sense its like a triangle then a segment

- jim_thompson5910

oops typo

- jim_thompson5910

let me fix

- anonymous

i see what you mean. oops. aha

- jim_thompson5910

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- anonymous

Intersecting Chords Theorem?

- jim_thompson5910

nope

- anonymous

so many theorems.

- jim_thompson5910

http://www.regentsprep.org/regents/math/geometry/gp11/LsimilarProof.htm

- jim_thompson5910

keep in mind that the 2 column proof builds up to what we want to aim for. Each step/line is necessary in getting to where we want to go

- anonymous

AA?

- jim_thompson5910

yep, Angle Angle
|dw:1433205268084:dw|

- jim_thompson5910

|dw:1433205298484:dw|

- anonymous

Ohh.

- jim_thompson5910

so we have a pair of congruent corresponding angles, therefore that's why \(\large \triangle ABC \sim \triangle DBA\) is true (because of the AA similarity theorem)

- anonymous

I see what you mean.

- anonymous

then what about the last one?

- jim_thompson5910

any thoughts on it?

- anonymous

I mean. I see how they relate. corresponding sides?

- jim_thompson5910

how can we use line 5 to lead up to line 6 ?

- anonymous

Honestly I have no idea.

- jim_thompson5910

if you look through these theorems
http://www.regentsprep.org/regents/math/geometry/gp11/LsimilarProof.htm
which theorem deals with proportions?

- anonymous

SSS!!!(:

- jim_thompson5910

yes specifically the converse of the SSS similarity theorem

- jim_thompson5910

the SSS similarity theorem
if the sides are all in proportion (as shown in the fractions), then the triangles are similar

- jim_thompson5910

converse of the SSS similarity theorem
if the triangles are similar, then the sides form a proportion

- jim_thompson5910

the converse is the "backwards" version of the original, so to speak

- anonymous

I see. so does this look good?

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- jim_thompson5910

it's not just SSS theorem, it's the converse of that theorem

- jim_thompson5910

everything else looks good though

- anonymous

so write converse sss theorem?

- jim_thompson5910

yeah or something like that so the teacher knows

- anonymous

BTW this isn't a test or quiz or anything.. I see alot of students ask for answers on here.

- jim_thompson5910

ok I hope that it's not. Those sorts of things should be taken individually without any help. If anything, they should be proctored.

- anonymous

Anyways Thank you for the help!! It means the world!

- jim_thompson5910

you're welcome

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