We use the same notation as in the statement of the theorem in the writeup on Pythagorean triples.

The proof is quite easy and it comes down to formulae that you might recognise from calculus for the sine, cosine and tangent of a double angle.

Start with the unit circle v2+w2=1 centred at the origin and with radius one. We can give a closed formula for a point (v,w) on the circle. A typical point is
(v,w)=(1-t2) / (1+t2), 2t / (1+t2)) (*).

The proof of this assertion is a consequence of those trigonometric identities I mentioned but a more geometric point is given as follows (you will want to draw a diagram to understand this). We project the circle onto the w-axis. Consider the line that goes through the point (-1,0) and hits the w-axis at the point (0,t). Where does this line meet the circle? Well it meets it exactly at the point (*) above.

It's clear that this point (v,w) is a pair of rational numbers if and only if t is rational. So to get back to the proof of the theorem, suppose that we have a primitive Pythagorean triple (a,b,c). Then (a/c,b/c) is a point on the unit circle. So by the above argument there corresponds a rational number t to this point. Write t=x/y in its lowest terms (so that x and y are coprime). The formula (*) nows gives us that a=x2-y2, b=2xy and c=x2+y2. This shows that every primitive triple has the required form. That triples of the form in the statement are primitive Pythagorean triples is clear, so we are done.