Consider a circle of diameter d. 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?