Prove the Converse of the Pythagorean theorem. The converse of the Pythagorean theorem states that when the sum of the squares of the lengths of the legs of the triangle equals the square length of the hypotenuse, the triangle is a right triangle. Be sure to create and name the appropriate geometric figures. This figure does not have to be submitted.

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.

A community for students.

Prove the Converse of the Pythagorean theorem. The converse of the Pythagorean theorem states that when the sum of the squares of the lengths of the legs of the triangle equals the square length of the hypotenuse, the triangle is a right triangle. Be sure to create and name the appropriate geometric figures. This figure does not have to be submitted.

Geometry
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free 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 this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

please help me ! :(
We are given \[\triangle ABC\] and \[a^2+b^2=c^2\] and have to prove that \[\angle ACB=90^{\circ}\] We begin by constructing another triangle, \[\triangle EFG\] right angled at G and having sides a and b(this is by construction, we have purposely drawn it's sides as a and b) applying pythagorus theorem in EFG we have \[n^2=a^2+b^2\] But we are given that \[c^2=a^2+b^2\] \[\implies c=n \implies AB=EF\] and \[AC=EG\](by construction) \[BC=FG\](by construction) thus we conclude that \[\triangle ABC \cong \triangle EFG\] (By SSS criterion) Thus ABC and EFG are congruent by SSS congruency criterion \[\implies \angle ACB=\angle EGF \space \space \space (cpct)\] thus angle acb and egf are equal because they are corresponding parts of congruent triangles and corresponding parts of congruent triangles are EQUAL but we know that angle EGF is 90 by construction \[\implies \angle ACB=90^{\circ}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1442687055553:dw|
omg thank you SO MUCH. this is the only question I was stuck on and also the last question for the entire class. you really just helped me a ton! @Nishant_Garg
You're welcome, I hope you understood the proof!
I did after reading over it twice and drawing it out :) lol @Nishant_Garg

Not the answer you are looking for?

Search for more explanations.

Ask your own question