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anonymous

  • one year ago

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.

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. anonymous
    • one year ago
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    please help me ! :(

  3. anonymous
    • one year ago
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    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 \[\implies \angle ACB=90^{\circ}\]

  4. anonymous
    • one year ago
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    |dw:1442687055553:dw|

  5. anonymous
    • one year ago
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    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

  6. anonymous
    • one year ago
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    You're welcome, I hope you understood the proof!

  7. anonymous
    • one year ago
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    I did after reading over it twice and drawing it out :) lol @Nishant_Garg

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