Geometry Proof (Circles and Chords)

Hey all. At my school we have these things called "Real Problems" that we get every 3-4 weeks. This time it's about proofs. I have an A- in the class, so I got the difficult one with 4 very difficult (IMO) proofs.

I have NO idea where to start on this..I have drawn it out for you. All help is greatly appreciated. I am here to merely learn, not copy everything. I would appreciate full answers, but I am definitely here to Learn how to do this for tests.

Given: Tangent line AB and chord BC.

Prove: <ABC = (1/2)(arc)BC

*In other words prove: **(Angle ABC = Half of arc BC)*

HINT: Using the center of the circle, draw in OB and OC (which I did in the picture for you).

http://img249.imageshack.us/img249/4413/proof1f.jpg

Re: Geometry Proof (Circles and Chords)

The question cannot possibly be right because the dimensions of an angle and arc-length are different.

Try using $\displaystyle s=r\theta$ for the *correct* version of the question.

Re: Geometry Proof (Circles and Chords)

1) How do we know this is not some sort of contest problem and our participation would be illegal?

2) How do we know this isn't an examination where outside help would be inappropriate?

3) If you really have NO IDEA, either you should not have been given the problem or the problem was intend to stretch you. In eaither case, outside influence will not accomplish the intent.

You start by gathering everything you know about chords and circles. There, now you know where to start.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**alexmahone** The question cannot possibly be right because the dimensions of an angle and arc-length are different.

Try using $\displaystyle s=r\theta$ for the *correct* version of the question.

The arc is in degrees as well as the angle. I'm unsure on where to start is all.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**TKHunny** 1) How do we know this is not some sort of contest problem and our participation would be illegal?

2) How do we know this isn't an examination where outside help would be inappropriate?

3) If you really have NO IDEA, either you should not have been given the problem or the problem was intend to stretch you. In eaither case, outside influence will not accomplish the intent.

You start by gathering everything you know about chords and circles. There, now you know where to start.

Lol dude..I'm in 10th grade just trying to get some math help. Nothing wrong about that.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Volux** The arc is in degrees as well as the angle. I'm unsure on where to start is all.

Refer to the *Alternate Segment Theorem*.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**alexmahone** Refer to the *Alternate Segment Theorem*.

I tried Googling this theorem but found nothing like mine. Is there an easy way to get around it..I just need to find out where to start..like the second step of the proof. I will be using a 2 column proof format.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Volux**

Note that $\displaystyle \Delta OBC$ is isosceles.

So $\displaystyle 2m\left( {\angle CBO} \right) + m\left( {\angle COB} \right) = 180^o$.

Also $\displaystyle m\left( {\angle CBO} \right) + m\left( {\angle CBA} \right) = 90^0 \;\& \,m\left( {\angle COB} \right) = m(\text{arc}(BC))$.

Put that together. The result 'falls out'.

Re: Geometry Proof (Circles and Chords)

So for the OBC triangle being isosceles the justification would be definition of an isosceles triangle?

Also,for 2<CBO + <COB = 180. and the other one, how do you know that? Is there a theorem or some justification for that? This would help me so I can fully understand and write the 2 column proof.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Volux** So for the OBC triangle being isosceles the justification would be definition of an isosceles triangle?

Also,for 2<CBO + <COB = 180. and the other one, how do you know that? Is there a theorem or some justification for that? This would help me so I can fully understand and write the 2 column proof.

What is in the water today? This is the second post in a row of people trying to do something without knowing the fundamentals.

Surely you can see that $\displaystyle m(\overline{OC})=m(\overline{OB})~?$

So the triangle is isosceles.

Surely you know the angle sum theorem?

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Plato** What is in the water today? This is the second post in a row of people trying to do something without knowing the fundamentals.

Surely you can see that $\displaystyle m(\overline{OC})=m(\overline{OB})~?$

So the triangle is isosceles.

Surely you know the angle sum theorem?

Lol yes I see it, but we have the have proof. And you said it, the angle sum theorem. I have it somewhere in my notes, it's just that I am not good with recognizing what is what. I have a hard time with proofs.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Volux** Lol yes I see it, but we have the have proof.

There is nothing there to prove.

Radial segments in a circle have the same length, the radius.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Plato** There is nothing there to prove.

Radial segments in a circle have the same length, the radius.

I'm to prove that angle ABC = half of arc BC though.

This is what my proof looks like right now:

1. Tangent line AB and chord BC 1. Given

2. Triangle CBO is isosceles 2. Angle sum theorem

3. 2<CBO + <COB = 180 3. ?

4. <CBO + <CBA = 90 4. ?

5. <COB = (arc)BC 5. ?

6. <ABC = (1/2)(arc)BC 6 ?

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**Volux** Lol dude..I'm in 10th grade just trying to get some math help. Nothing wrong about that.

You failed to answer the "How do we know" parts.

Re: Geometry Proof (Circles and Chords)

Quote:

Originally Posted by

**TKHunny** You failed to answer the "How do we know" parts.

I am unsure of how to prove this as well..