Prove SSS theorem.
Please, help.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- Loser66

Prove SSS theorem.
Please, help.

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- Loser66

- Loser66

- jackthegreatest

I can help but wats the problem?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Loser66

Prove SSS theorem, that is the problem.

- jackthegreatest

Oh I thought u had to prove two triangles similar using sss

- Loser66

\(\triangle ABC \) and \(\triangle A'B'C'\) such that AB = A'B', AC = A'C', BC = B'C'.
Prove that \(\triangle ABC \cong \triangle A'B'C'\)

- jackthegreatest

I can't help on proving theorem itself sorry

- jackthegreatest

Hmmm

- Loser66

It's ok, friend. Thanks for being here.

- jackthegreatest

Sorry can't help

- anonymous

SSS is a postulate

- zzr0ck3r

right, I don't think you prove this one.
It would be like proving The Axiom of Choice without one of its equivalents.

- Loser66

Thanks for replying, but yes, we are and I have it done.

- zzr0ck3r

ahh, I never took geometry but when I googled it, it seemed like it was not something you prove.

- anonymous

SSS is a postulate in the original formulation of Euclidean geometry: https://en.wikipedia.org/wiki/SSS_postulate
depending on what alternative axiomatization you use for your modern geometry course, sure, you can prove it from that alternative foundation, but without stating that before it is only natural to assume we're talking about classical Euclidean geometry. it is impossible to even answer your question without your axiomatization anyways so this question as it stands is ill-posed and impossible to answer

Looking for something else?

Not the answer you are looking for? Search for more explanations.