# Pythagoras' from Broken Chords

Bui Quang Tuan found a way to derive the Pythagorean Theorem from the Broken Chord Theorem.

Let ABC be a triangle with a right angle at C, AB is diameter of circumcircle (O). If P is the midpoint of the arc ACB, the APB is right isosceles triangle, with the right angle at P. Assuming AC > BC, let M be the orthogonal projection of P on AC. Line PM cuts AB at N, B' is orthogonal projection of B on PM, as in the diagram:

As usual, let a = BC, b = AC, c = AB.

Since BCMB' is rectangle,

 (1) MB' = CB = a.

By the Broken Chord Theorem,

 (2) AM = (a + b)/2 (3) MC = (a + b)/2 - a = (b - a)/2

And, since PMC is right isosceles triangle at M,

 (4) MP = MC = (b - a)/2.

Further,

 (5) Area(MNB) = Area(ANB')

because AM||B'B.

Now, the area of APB equals c²/4. We find it yet in a different way:

 c²/4 = Area(APB) = Area(APM) + Area(PMB) + Area(AMN) + Area(MNB) = Area(APM) + Area(PMB) + Area(AMN) + Area(ANB') = Area(APM) + Area(PMB) + Area(AMB').

Using (1), (2), (3), (4) to calculate areas of triangles, we obtain:

 Area(APB) = MP·AM/2 + MP·MC/2 + AM·MB'/2 = ((b - a)/2)·((a + b)/2)/2 + ((b - a)/2)·((b - a)/2)/2 + ((a + b)/2)·a/2 = (a² + b²)/4

This means c²/4 = (a² + b²)/4 or c² = a² + b². The proof is complete.