# 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

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 | |

= (a² + b²)/4 |

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

|Contact| |Front page| |Contents| |Geometry| |Pythagorean Theorem|

Copyright © 1996-2018 Alexander Bogomolny