Pythagoras’ theorem is one of the central results of Euclidean geometry. In any right triangle with legs
\(a\) and \(b\) and hypotenuse \(c\), the areas of the squares built on the legs add up
exactly to the area of the square built on the hypotenuse:
\[
a^2+b^2=c^2.
\]
Although this formula is often memorized, its meaning is geometric: it connects lengths to areas.
A proof becomes easier to trust when you can match every algebraic step with a piece of the diagram.
A classic visual proof (often attributed to van Schooten) is a rearrangement argument. Start with a large square
whose side length is \(a+b\), so its total area is \((a+b)^2\). Inside that square, place four congruent right
triangles with legs \(a\) and \(b\). Each triangle has area \(\tfrac{ab}{2}\), so the four triangles
together cover \(4\cdot\tfrac{ab}{2}=2ab\) units of area. Since the big square’s area is fixed, the leftover region
(big square minus the four triangles) must have area
\[
(a+b)^2-2ab=a^2+b^2.
\]
Now look at how the leftover region changes with placement. In the “pinwheel” placement, the hypotenuses of the four triangles
outline a tilted inner square. That inner square has side length \(c\) (the hypotenuse of the original triangle), so its
area is \(c^2\). In a second placement, the same four triangles are slid into two rectangles, and the leftover region becomes
two axis-aligned squares of areas \(a^2\) and \(b^2\). The outer square and the triangles are unchanged—only the
arrangement changes—so the leftover area must be the same in both pictures. Therefore,
\[
c^2=a^2+b^2.
\]
This is why rearrangement proofs feel so satisfying: the equality is literally an “area conservation” statement.
There is also a famous proof based on similar triangles, which explains the theorem through proportionality.
Drop an altitude from the right angle to the hypotenuse, meeting it at a point \(H\) and splitting the hypotenuse into
two segments \(p\) and \(q\). The two smaller triangles are similar to the original right triangle. From similarity
you can show
\[
a^2=c\,p,\qquad b^2=c\,q,
\]
and because the two segments make the whole hypotenuse, \(p+q=c\). Adding the two relations gives
\[
a^2+b^2=c(p+q)=c\cdot c=c^2.
\]
So even without squares, the theorem follows from a scaling relationship hidden inside the right triangle.
In applications, Pythagoras lets you compute an unknown side when the other two are known. The 3-4-5 triangle is the most common
check: \(3^2+4^2=9+16=25=5^2\). Use the calculator’s animation to connect each proof step to a visible piece of geometry,
not just symbols on a page.
