- #1
prane
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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?