# Circle Geometry Theorem

• Jan 23rd 2009, 07:27 PM
xwrathbringerx
Circle Geometry Theorem
Hi

I was just wondering...

http://www.mathhelpforum.com/math-he...ure361-cg1.bmp

For a question, can I say angles BAO and OCD are equal (angles at the circumference standing on the same arc BD are equal) or is that wrong?

I tried to prove the two triangles congruent in order to prove the angles were equal but failed miserably. (Crying)

Thanx
• Jan 23rd 2009, 10:23 PM
mollymcf2009
For a question, can I say angles BAO and OCD are equal (angles at the circumference standing on the same arc BD are equal) or is that wrong?

When you refer to the angles on the arc BD being equal are you referring to the angles BAO & OCD? Because those angles aren't on the arc BD. The only angle "on" arc BD is the central angle BOD. Were you given that any angles are equal? **Remember that when you refer to an angle, the middle letter needs to be the vertex of the angle*

I tried to prove the two triangles congruent in order to prove the angles were equal but failed miserably. Even if they were congruent, it wouldn't be based on those two angles being equal because they aren't the same angle on each triangle, if you rotate them around so they are on top of eachother, you will see what I mean.
Can you give me the original question provided? There really isn't enough information here to figure that out.
• Jan 24th 2009, 01:27 AM
xwrathbringerx
Sori the original question was:

AB and CD are two perpendicular chords of a circle, centre O. Prove that angle BOC + angle AOD = 180 degrees.
• Jan 24th 2009, 06:05 AM
Angles in circles
Hello xwrathbringerx
Quote:

Originally Posted by xwrathbringerx
Sori the original question was:

AB and CD are two perpendicular chords of a circle, centre O. Prove that angle BOC + angle AOD = 180 degrees.

Suppose that AB and CD meet at X.

Angle BOC = 2 x angle BAC (Angle at centre = twice angle at circumference, when subtended by the same arc) (1)

Angle AOD = 2 x angle ACD (Angle at centre) (2)

In triangle ACX, angle AXC = $\displaystyle 90^o$ (Given)

$\displaystyle \Rightarrow$ angle XAC + angle XCA = $\displaystyle 90^o$ (Angle sum of triangle)

$\displaystyle \Rightarrow$ angle BOC + angle AOD = $\displaystyle 180^o$, from (1) and (2)

Is that OK now?

• Jan 24th 2009, 06:35 AM
Plato
Here is yet another way.
It follows from the given $\displaystyle \overline {AB} \bot \overline {CD}$ that $\displaystyle m\left( \text{arc(AC)} \right) + m\left( \text{arc(BD)} \right) = 180$.
If we remove those two arcs from the whole, the we get arcs subtended by the angles in question.
Therefore their sum is also 180.
• Jan 24th 2009, 06:46 AM
mollymcf2009
AB and CD are two perpendicular chords of a circle, centre O. Prove that angle BOC + angle AOD = 180 degrees.

Two perpendicular chords create four 90 degree angles at their intersection.

The diameter of a circle is the largest chord in a circle.

Vertical angles of two perpendicular lines are congruent.

Chord AB & chord CD are perpendicular. At their intersection, the angles AOD, AOC, BOC & DOB are created. Point O is the center of the circle. Angles AOD, AOC, BOC & DOB are all 90 degree angles. Angles BOC & AOD are vertical angles. Angle BOC + angle AOD = 180 degrees