The converse of Pythagoras

Can you prove that triangles are right-angled when $a^2+b^2=c^2$?

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Problem

Pythagoras' theorem states that:

"If a triangle with sides $a, b, c$ has a right-angle, and $c$ is the hypotenuse, then $a^2+b^2=c^2$."

You can find some proofs of Pythagoras' theorem here and here For this problem you can assume that Pythagoras' theorem is true.

The converse of Pythagoras' theorem states that:

"If a triangle has lengths $a, b$ and $c$ which satisfy $a^2+b^2=c^2$ then it is a right-angled triangle."

We can prove this by using a proof by contradiction, which means starting by assuming that the opposite is true and then showing that this leads to an inconsistency or contradiction.

Below is a proof sorter of a proof that if $a^2+b^2=c^2$ then $C$ cannot be less than $90^{\circ}$, and a diagram to accompany the proof.  Can you use the diagram and the statements to reconstruct a proof that angle $C$ cannot be less than $90^{\circ}$?

Image
The Converse of Pythagoras

 



Can you use the diagram below, and the ideas from the proof sorter above, to construct a proof that angle $C$ cannot be more than $90^{\circ}$?

Image
The Converse of Pythagoras

Once both proofs have been completed then we know that $C$ cannot be less than $90^{\circ}$, and that $C$ cannot be more than $90^{\circ}$, therefore $C$ must be equal to $90^{\circ}$ and hence if we have $a^2+b^2=c^2$ then the triangle is right-angled at $C$.

 

We are very grateful to the Heilbronn Institute for Mathematical Research for their generous support for the development of this resource.