# Urgent Circle Proof

• Sep 7th 2007, 06:34 PM
trent19
Urgent Circle Proof
I need to prove that a radius of a circle bisects a chord at 90 degrees, using a vector method and another method. I bet the proof is simple but it has confused me, seeming as I have always taken this as a property. If you can help, thanks heaps.
• Sep 7th 2007, 06:46 PM
Krizalid
A little picture

http://img525.imageshack.us/img525/8528/sdfsdjj4.png

We consider a circle centered at $\displaystyle O$

All you need to show is that $\displaystyle \triangle OCB\cong\triangle OCA$ and you are done.
• Sep 7th 2007, 07:07 PM
trent19
Thanks heaps, should be right now. Have a good one
• Sep 7th 2007, 11:52 PM
trent19
Sorry, but maybe i wasn't right. I have stared at that for hours, and copied it repeatedly but i still can't work out how to show that the triangles are congruent. If anyone can help with the next step.

That and any ideas on how to solve it using vectors. I know it has something to do with dot product (if u.v=0, u perp to v) but cant see how to get to that point.

Once again sorry, but if anyone has any ideas or can show me what to do, i will really appreciate it.

Thanks

P.S, sorry about the double post. I am new at this and wasnt sure how else to get this message out.
• Sep 8th 2007, 03:37 AM
ticbol
Quote:

Originally Posted by trent19
I need to prove that a radius of a circle bisects a chord at 90 degrees, using a vector method and another method. I bet the proof is simple but it has confused me, seeming as I have always taken this as a property. If you can help, thanks heaps.

This not by way of vectors.

Let each half of the chord is c/2.
And segment of bisecting radius from center to midpoint of chord be b.

Two radii to both ends of the chord, and the chord itself, make an isosceles triangle, with the cchord as the base. So the base angles are congruent.

Are the two angles separated by b at the apex of the isosceles triangle equal?
Say one of them is angle alpha, the other, angle beta.
---cos(alpha) = b/r
---cos(beta) = b/r also
So alpha = beta
Hence the apex angle of the isosceles triangle is bisected by b also

Now, the two base angles are equal.
The two acute angles forming the apex of the isosceles triangle are equal.
Therefore the third angles of the two triangles formed by b on the isosceles triangle are equal also.
If these two third angles are equal, then each measures 90 degrees, because these two form a straight angle or 180 degrees.

That's it.

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Edit:

Are the two angles separated by b at the apex of the isosceles triangle equal?
Say one of them is angle alpha, the other, angle beta.
---cos(alpha) = b/r
---cos(beta) = b/r also
So alpha = beta

That is very weak. It is not established yet that the two triangles divided by b are right triangles.

This is better:
Comparing the two triangles,
---(side r of one) = (side r of the other)
---(side c/2 of one) = (side c/2 of the other)
---(side b of one) = (side b of the other)
Therefore the two triangles are congruent. By SSS.
Therefore, alpha = beta. By corresponding angles.
• Sep 8th 2007, 03:56 AM
trent19
Thanks HEAPS for that. It helped... A LOT!
• Sep 8th 2007, 12:20 PM
Krizalid
Actually, it's not necessary to apply trig. for this.

We have that $\displaystyle \triangle OCB\cong\triangle OCA$ by ASA congruence.
• $\displaystyle \overline{OA}=\overline{OB}$ (radii)
• $\displaystyle \overline{OC}=\overline{OC}$ (common side between $\displaystyle \triangle OCB$ & $\displaystyle \triangle OCA$)
• Therefore $\displaystyle \triangle OCB\cong\triangle OCA$ $\displaystyle \because\measuredangle\,COA=\measuredangle\,COB$
• It follows that $\displaystyle \overline{CA}=\overline{CB}\,\blacksquare$
• Sep 8th 2007, 01:40 PM
topsquark
Quote:

Originally Posted by Krizalid
Actually, it's not necessary to apply trig. for this.

We have that $\displaystyle \triangle OCB\cong\triangle OCA$ by ASA congruence.
• $\displaystyle \overline{OA}=\overline{OB}$ (radii)
• $\displaystyle \overline{OC}=\overline{OC}$ (common side between $\displaystyle \triangle OCB$ & $\displaystyle \triangle OCA$)
• Therefore $\displaystyle \triangle OCB\cong\triangle OCA$ $\displaystyle \because\measuredangle\,COA=\measuredangle\,COB$
• It follows that $\displaystyle \overline{CA}=\overline{CB}\,\blacksquare$

Actually since you are trying to prove that angles OCA and OCB are right, you are probably better off noting that:
1) OA = OB (both are radii of the circle)
2) OC = OC (reflexive property)
3) AC = BC (given, because the radius bisects the chord)

So triangles OAC and OBC are congruent by SSS. Thus angles OCA and OCB are equal. But they are also supplementary. Thus OCA = OCB = 90.

-Dan
• Sep 8th 2007, 01:43 PM
Krizalid
Quote:

Originally Posted by trent19
I need to prove that a radius of a circle bisects a chord at 90 degrees

This is the hypotesis.

I wasn't proving that those angles are right.
• Sep 12th 2007, 05:02 AM
trent19
Thanks heaps. It helped a lot. I really had no idea but it turned out to be relatively simple. :) Thanks again