Sunday, September 14, 2008

Circular Argument I

Here's something that caught my attention when I was tutoring one of my students in 'E' Math at the Secondary School level: the angle subtended by a chord of a circle at the centre of the circle is twice that subtended by the same chord at the circumference.

Perhaps a diagram might make things much easier:


The chord is the line AB, and the angle subtended at the centre is the angle AOB, and the angle subtended at the circumference is the angle ACB, as indicated in Greek above, haha. So this property of a circle says that angle AOB is twice that of angle ACB. But why?

Easy enough, I shall use the method that Shaun used during our lecture, which is a rather neat and easy proof; so kudos to Shaun! First of all, you divide the triangle ABC into half, down the centre as shown below with a line CD, and then notice I've coloured an isoceles triangle in green:

And then you should notice that this line should bisect the angles AOB and ACB, and we can conclude first that:


And from this, the following statements should therefore make sense:


Quite a neat proof huh? Shaun came up with it all by himself, pro!

1 comment:

quirK said...

I didn't come up with it...