Are all triangles where "$a^2 = b^2+ c^2$", right-angled? For a right angle triangle, you can say that the square of the hypotenuse is equal to the sum of the squares of the other two sides.  Does the converse hold, ie. can you also say that, if the square of the hypotenuse is equal to the sum of the squares of the other two sides, then the triangle MUST be right-angled?
 A: The word hypotenuse applies only to right triangles in the first place, but we can just as well ask if for a triangle with sides $a, b, c$ whether satisfying the Pythagorean Identity $a^2 + b^2 = c^2$ implies that the angle between $a$ and $b$ is right.
The answer is yes; this follows from (for example) the Law of Cosines, which generalizes the Pythagorean Theorem. If a triangle has sides $a, b, c$ and the angle opposite $c$ is $\gamma$ then
$$c^2 = a^2 + b^2 - 2 a b \cos \gamma .$$
Then, the Pythagorean Identity implies that $\cos \gamma = 0$, or $\gamma = \frac{\pi}{2}$.
A: Yes it does by Cosine Rule:
$$\cos A=\frac{a^2-b^2-c^2}{2bc}=0=\cos\left(\frac{\pi}2\right)$$
Does that imply $\displaystyle A=\frac{\pi}2$ unless $A=3\pi/2,5\pi/2,\cdots$ or $A=-\pi/2,-3\pi/2,\cdots$ both of which never hold.
A: The law of cosines:
$$c^2=a^2+b^2-2ab\cos{C}$$
If $2ab\cos{C}=0$, where $C$ is the angle opposite side $c$, what can you tell me about $\cos{C}$?
A: Law of cosine: $a^2=b^2+c^2+2 a b \cos{(A)}$
Knowing that $a^2=b^2+c^2$ we have that $2 a b \cos{(A)}=0 $
Since $a$ and $b$ are not equal to $0$, it should be $\cos{(A)}=0 \Rightarrow A=\frac{\pi}{2}$
A: The converse does hold. This is Proposition 48, Book I of Euclid's Elements. (The Pythagorean Theorem is I-47.)
I recommend looking up Euclid's proof, since it makes nice use of geometric inequalities, and as such is subtler than the proof of I-47. The edition linked to also happens to be very attractive!
