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### Inscribed Angle Theorem

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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?
```

```
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

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