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Simple geometric proof

  1. Aug 20, 2012 #1
    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?
  2. jcsd
  3. Aug 20, 2012 #2
    What are you getting at?

    Only right angled triangles have a hypotenuse. So by definition your triangle is right angled.

    However you can easily inscribe a non right angled triangle (that does not therefore have a hypotenuse).

    What you are trying to describe and prove is more conventionally described as the angle on a semicircle - which is always a right angle.

    The actual proof you use must depend upon the way your geometry has built up ie what theorems have already been proved and are therefore available for use.

    The original proof used by Euclid, who was not into trogonometry, was based on the theorem that the angle subtended by a chord to the centre is twice the angle subtended to the circumference. Do you have this theorem?
    Last edited: Aug 20, 2012
  4. Aug 20, 2012 #3


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    One basic property of angles and circles is that the measure of an angle inscribed in a circle is 1/2 the measure of the arc it cuts off. If one side of an inscribed triangle is a diameter of the cirle, then the opposite angle cuts off an arc of 180 degrees and so its measure is 90 degrees.
  5. Aug 21, 2012 #4

    What do you mean by the measure of an angle? I don't really follow your argument
  6. Aug 21, 2012 #5
    That's just a modern posh way to say what Euclid said.
  7. Aug 21, 2012 #6
  8. Aug 22, 2012 #7


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    Proving your special case of the general theorem is easy. With your notation of A B and C, let the center of the circle be O.

    Triangle AOB is isosceles. Let angle OAB = angle OBA = x.
    Triangle BOC is also isosceles. Let angle OBC = angle OCB = y.
    Angle ABC = x + y.
    The sum of the angles of triangle ABC = 2x + 2y = 180 degrees, so x + y = 90.
  9. Aug 22, 2012 #8
  10. Aug 22, 2012 #9
    That's a really nice proof! I didn't realize there was an alternative to the one Hallsofivy cited.
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