Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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/   
    
Associated Topics:
High School Conic Sections/Circles
High School Euclidean/Plane Geometry
High School Geometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/