Draw a circle. Now pick two points (A,B) on the periphery of that

[kenewbie]

Draw a circle. Now pick two points (A,B) on the periphery of that circle, so that a line between A and B will have to pass through the center of the circle. The line AB is, in other words, the diameter of the circle.

Now, pick a point C anywhere on the periphery of the circle. Prove that in any such triangle ABC, angle C is 90 degrees.

--

I have no idea. At first I was even reluctant to agree that C is always 90 degrees, but making random points on a circle seem to confirm this.

The best I can come up with is that if I draw a normal from point C to AB, then that normal will always make 90 degree angles, so both "split-triangles" must be similar and so the original ABC must be similar as well, and contain a 90 degree angle.

I don't think that constitutes a proof.

 

[rock.freak667]

Put O as the centre of the circle and draw in the line OC. Noting that OC is a radius,as well as OA and OB. Put angle OAC=a and OBC=b. Now just use some simple triangle properties and you should get it.

 

[kenewbie]

Put O as the centre of the circle and draw in the line OC. Noting that OC is a radius,as well as OA and OB. Put angle OAC=a and OBC=b. Now just use some simple triangle properties and you should get it.

I think that since OC and AC are both R long, angle ACO must be equal to angle A.
And for the same reason, angle BCO must equal B.

Which means angle C should be A+B.

A+B+C = 180
A+B+(A+B) = 180
2(A+B) = 180
A+B = 90
C = 90

That seems to work. Thanks a lot rock.freak, I don't think I would ever have thought to split the triangle between C and O, since the new ones seem so random :)

k

 

[HallsofIvy]

There is a fairly well know theorem that an angle with vertex lying on a circle has angle equal to 1/2 the arc it subtends. If the two points are ends of a diameter, then the arc is subtends is 1/2 a circle, 180 degrees, so the angle is 90 degrees.

 

[tiny-tim]

There is a fairly well know theorem that an angle with vertex lying on a circle has angle equal to 1/2 the arc it subtends. If the two points are ends of a diameter, then the arc is subtends is 1/2 a circle, 180 degrees, so the angle is 90 degrees.

Hi kenewbie and HallsofIvy!

Yes … the method of proof is exactly the same, plus exterior angle of triangle = sum of the two opposite angles.

 

[kenewbie]

Ivy, ok. I've never learned that, perhaps it comes later.

Now I know why :)

k