Geometry Proof (Circles and Chords)

[XcapeX]

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).

 

[Deleted member 7467]

I remember doing this last year but don't remember how to do them. Sorry bud

 

[Tressy]

Hmm, I hope I don't sound stupid asking this. Pardon me if I do, though. But how can an angle be equal to half a length? I mean, they're not even the same thing. If I'm not wrong, arc means a portion of the circumference. Unless you mean to prove that angle ABC = 1/2 angle BOC? Maybe you guys just use the terms differently. 😛

Anyway, if it's really asking you to prove that <ABC = 1/2 (<BOC), hopefully this will help.



And if you want the solution written out,

Spoiler

<ABO = 90 degrees because it's tangent to the circle at B, so OB and AB are perpendicular.

And since, OB = OC = radius of circle, triangle BOC is an isosceles triangle.

So let's say <OBC = x degrees,
<OBC = <OCB = x degrees as they are angles of the isosceles triangle

Therefore, you can deduce that <BOC = 180-2x degrees (angles in triangle add up to 180 degrees), and 1/2 (<BOC) = 90-x degrees

And also, <ABC = 90-x degrees (<ABO is right-angled)

Therefore, <ABC = 1/2(<BOC) = 90-x degrees 🙂

But if what I thought it means isn't what it really is asking you to prove, then I just wasted my time typing all that out. LOL.