- #1

- 23

- 0

## Main Question or Discussion Point

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?

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?