Consider a circle of diameter d.(adsbygoogle = window.adsbygoogle || []).push({});

Inscribe a triangle within the circle so that the triangle has hypotenuse d.

Prove that this triangle is always right angled.

If we define the 3 points on the circumference of the circle that define the triangle as A B and C such that |AC| = d then we have AB + BC = AC

Also define the angle at C to be σ

The right angle is at point B thus we need

(AB).(BC) = 0

now AB = AC - BC

thus (AB).(BC) = (AC - BC).(BC) = (AC).(BC) - |BC|^2 = |AC||BC|cos(σ) - |BC|^2

= |BC|(|AC|cos(σ) - |BC|)

now this is equal to 0 if and only if |BC| = |AC|cos(σ) however this is only true if and only if ABC is a right angled triangle so I haven't really proved it :/

any suggestions?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Simple geometric proof

Loading...

Similar Threads - Simple geometric proof | Date |
---|---|

A Simple metric tensor question | Aug 14, 2017 |

Simple/Basic Example on Wedge Products | Feb 4, 2016 |

Simple(?) index notation | Oct 22, 2015 |

Is every smooth simple closed curve a smooth embedding of the circle? | May 8, 2014 |

**Physics Forums - The Fusion of Science and Community**