## The circumcircle and the law of sines

The other day I posted my solution to a lovely puzzle I’d seen that Ed Southall (@solvemymaths) had come up with, I assumed there would be better more concise solutions and ask for people to send me their solutions. I got a number of them, including this one from Xxxygorzao (@toodrunkforzero) :

At first one of the steps wasn’t obvious to me. It was the step that uses a / Sin A = 2r. I questioned Xxxygorzao who told me that it was a property of all triangles that the ratio of side / opposite angles is equal to the diameter of the circumcircle. I thought this was a rule I had forgotten, but after thinking about it I’m not entirely sure if it is or if it’s one I have never encountered before. Xxxygorzao offered to link me to the proof of this theorem, but I thought it would be nice to try and find one myself as it didn’t jump out as self-evident that it was true.

I started with a circle, and drew a triangle with 3 points on the circumference (n.b. I used a pair of compasses to draw the circle, but this was just because I’m rubbish at sketching circles, it is just for sketch purposes.):

I labelled the points A, B and C, and intend to use the notation A, B and C for the angles at each point and a, b and c for their opposite sides as is conventional in the law of sines. I then sketched chord bisectors and labelled the midpoints D, E and F. Then I labelled their intersection point, and hence the centre of the circle O.

I considered the triangle A, D, O. I know that angle ADO is a right angle and that AD is half of b, as it’s a perpendicular bisector. I also know that the hypotenuse is a radius.

I looked back at my circle and realised that by inscribed angle theorem AOC is equal to 2 x ABC, or 2 x B in this case.

AOC is an isosceles triangle made from 2 radii, and thus the perpendicular bisector of the base is also an angle bisector, giving angle AOD to be equal to B.

The rest then followed from right-angled trigonometry.

It’s a nice fact to know, and I found justifying it to myself this way was both fun and allowed be to gain an insight and an understanding into why this fact is. Obviously, I’ve only shown it for the ratio with B and b here, but the others can be shown in a similar way, and they don’t need to be as they follow from the sine rule.

No need for law of sines. If O is the centre of the circle then angle BOM is a right angle (angle at the centre is twice the angle at the circumference theorem) and the solution is an elementary Pythagoras problem.

No need for law of sines to prove that the ratio given by the law of sines is equal to the diameter of a circumcircle?

Ah, do you mean to answer the question discussed previously which the answer that inspired the post was to? Aye, lots of ways to do it. I think I may write a follow up looking at some more.