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:

pythagorean theorem from broken chord theorem

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.

The Broken Chord Theorem

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