# Angles in a circle

• May 31st 2006, 05:48 PM
Vashti
Angles in a circle
http://s14.photobucket.com/albums/a3...rrent=scan.jpg

How would I prove that angle 2 is twice the size of angle 1?

thankyou ^^
• May 31st 2006, 06:13 PM
malaygoel
Quote:

Originally Posted by Vashti
http://s14.photobucket.com/albums/a3...rrent=scan.jpg

How would I prove that angle 2 is twice the size of angle 1?

thankyou ^^

firstly we will label the diagram
let the vertex with angle 1 be A
the two other unlabelled points be B and C
join OA and extend it
if you could follow the instructions upto here, then post a new diagram and I will give you the rest of the proof :)
• May 31st 2006, 06:14 PM
ThePerfectHacker
It is challenging to write out the proof without a diagram by the basic idea is this: If one of the lines that form that inscribed angle passes though the diameter then it clearly is true, why? Because the central angle is the exterior angle of the triangle which you have. But by the theorem: an exterior angle is the the sum of the non-abjacent angles of the triangle. Since the two non-adjacent angle are the same, because you have an isoseles. You can conlude it is twice as large as the inscribed angle.

In the case why the center does not line on one of those line, unlike in first paragraph. You need to break the problem as a sum of two arc (if center is included between the lines) and difference of big arc and small arc (if center is not included between the arcs).

Hope this helps.
----
Curious?
Why did you name yourself after an evil charachter in,
Esther from the Bible?
• May 31st 2006, 06:17 PM
Vashti
http://s14.photobucket.com/albums/a3...t=scan0001.jpg

like this?
(thankyou very much ^^)
• Jun 1st 2006, 05:17 AM
earboth
Quote:

Originally Posted by Vashti

Hello,

I've attached a diagram to explain how I labeled vertices, angles, etc.

The central line CO generates 2 isosceles triangles: AOC and BOC. I've marked the angles at the base of each triangle.

The angle $\displaystyle \angle(KOB) = \angle{(c_1)}+\angle{(b_1)}$ and$\displaystyle \angle(KOA) = \angle{(c_2)}+\angle{(a_2)}$. Thus

$\displaystyle \angle(AOB) = \angle{(c_1)}+\angle{(b_1)}+\angle{(c_2)}+\angle{( a_2)}=2 \cdot \angle(ACB)$

quod era demonstrandum

Greetings

EB