Inscribed Angle Theorem
Date: 01/21/99 at 09:17:57 From: Darren Subject: Circle Geometry - Inscribed Angle Theorem Hi, I'm trying to find a proof for the theorem stating that any angle theta inscribed on the same arc as a central angle is one half that central angle, where theta begins on the circle's circumference. I know you already showed someone how to look for a proof for the theorem, but I had trouble following it. (See http://mathforum.org/dr.math/problems/daniel2.12.96.html ) Dr. Byron said a proof could be initiated for the theorem by drawing an arc AB, a center C, radii AC and BC, vertex I on the circle, lines IA and IB, and finally the circle's diameter going through I and C. I was supposed to get two congruent isoceles triangles. Here's the problem. Dr. Byron claimed that AB, IC, and CB were radii (I assume instead of AB he meant CA, because AB is not always equal to the radius). I get isoceles triangles CIB and CAB, plus one other triangle I assume I do not need. The problem is, if these triangles are supposed to be congruent, how do I prove that IB = AB, even though I know all the other sides are radii? Do they not depend on angles ICB and ACB? Please help!
Date: 01/21/99 at 10:24:55 From: Doctor Floor Subject: Re: Circle Geometry - Inscribed Angle Theorem Hi Darren, Thank you for sending your question to Dr. Math! I think you misunderstood what Doctor Byron meant in his answer in the archives at: http://mathforum.org/dr.math/problems/daniel2.12.96.html Let me try to explain how you could prove the theorem. Consider a circle centered at C. Let DA be a diameter of this circle and let B be another point on the circle. You want to show that the angle on arc AB from C is double the angle on arc AB from D. Here is a picture for your reference: We see that angle(ACB) + angle(BCD) = 180 degrees, angle(BCD) + angle(DBC) + angle(CDB) = 180 degrees, and triangle BDC is isosceles, so we can conclude that angle(ACB) = 2*angle(CDB) = 2*angle(ADB). This proves your theorem for angles with one leg passing through the center of the circle. Other cases can be proved by adding a diameter to the picture. You can apply the simpler form twice in that new situation. Either adding or subtracting these two gives the proof. As an example consider the situation as in the following picture: For the theorem we should have to prove that angle(BCA) = 2*angle(BDA). In this picture the simpler form shows us that: angle(ECB) = 2*angle(EDB) angle(ECA) = 2*angle(EDA) But then also angle(BCA) = angle(ECA)-angle(ECB) = 2*(angle(EDA)-angle(EDB) = 2*angle(BDA) I hope this clears up what Doctor Byron meant. If you have a math question again, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum