 | Proofs for the Circle Theorems |  |
| Proofs for the Circle Theorems | |
When I did GCSE maths (and if you did it today) I learnt three 'circle theorems' which we used to find angles composed of circles, tangents and chords. Those theorems were never proved; a shocking state of affairs that failed to disturb me at the time. Now some time later I came across them again and wanted proofs. Unable to find proofs on the internet, I discovered proofs for myself. Sadly these proofs use a little A level technology (and certainly A level thinking), but I shall try to prove anything beyond GCSE that I use, so that if you have just been shown these as a GCSE student you can find out where they come from and why they're true.
I shall need some terminology in order to state the theorems, so I suppose I had better define those terms.
a line joining two points on the circles circumference.
given an angle between two lines inside the circle if we extend those lines till they meet the circle then take a chord joining them to form a triangle, then that chord is said to subtend the angle (or the angle is subtended by the chord).
actually this one's quite hard. Formally it's a line touching the circle with gradient equal to the circle's at that point, but if you've only done GCSE you likely don't know what that means. So for these purposes I will say it is a line that touches the circle such that it is at right angles to a radius drawn to the point of contact. (It turns out that any line which touches a circle just once satisfies this, but that's another matter).
It is also worth noting that I shall be working in degrees for these proofs. Although radians are vastly superior, GCSE students won't have met them.
Given a chord all angles subtended by that chord are equal.
For any chord the angle subtended at the centre is twice that subtended at the perimeter.
A tangent meets a chord at an angle equal to that subtended by the chord.
This is the hardest part: the first thing to realise is that up to the rotational symmetry of the circle any chord can be defined by the angle it subtends at the centre, say 2α (I've said 2α instead of α just to make the numbers a bit easier). Now any angle subtended by the chord is determined by a triangle of chords. Hence it is determined by picking one other chord sharing one end point with our base chord. Hence any particular choice of subtended angle may be given by choosing the angle subtended at the centre by our new chord, let us call this angle 2β 
. What does that mean? Well it means that if we can prove θ is independant of β then we'll have theorem 1: we'll know that it doesn't matter which of the angles subtended by a chord you pick, they're all equal.
To show that we're going to need to know how long the sides of our triangle are. In all of this the radius is just a factor of scaling, it doesn't make any difference, so we'll just say it's 1 to make it go away (you can do it with r instead and see that it all cancels if you like). Then we can find the length from the central angle by using the knowledge that sin gives us the opposite over the hypotenuse (the hypotenuse in this case is the radius, so it's 1). 
So the length of a chord with central angle 2α is 2 sin α. So we now have a triangle where we know the length of the three sides, and want to know one of the angles; enter the cosine rule!
The Cosine Rule: 
a2 = b2 + c2 - 2bc cos θ
So, let's plug it in and solve:
(2 sin α)2 = (2 sin β)2 + (2 sin (α + β))2 - 2(2 sin β)(2 sin (α + β))cos θ
sin2 α = sin2 β + sin2 (α + β) - 2 sin β sin (α + β)cos θ (divide by 4)
In order to proceed we need the expansion rule for sin(α + β):
sin(α + β) = sin α cos β + sin β cos α
This is a standard result, you can find it in any A level textbook. A rather good derivation is offered here.
We'll also need the relation:
sin2 θ + cos2 θ = 1
Another standard result, which I've have given a proof of.
At this stage all that remains is some algebra, it's probably best to just do it yourself on a scrap of paper. In case you get stuck I'll do it below:
Expanding sin2 (α + β) = (sin α cos β + sin β cos α)2
Expanding sin2 (α + β) = sin2 α cos2 β + 2 sin α cos β sin β cos α + sin2 β cos2 α
and substituting we get:
sin2 α = sin2 β + sin2 α cos2 β + 2 sin α cos β sin β cos α + sin2 β cos2 α - 2 sin β sin (α + β)cos θ
sin2 β - sin2 α + sin2 α cos2 β + 2 sin α cos β sin β cos α + sin2 β cos2 α = 2 sin β sin (α + β)cos θ
Now we need the relation: sin2 β - sin2 α + sin2 α cos2 β = sin2 β cos2 α
Pf of Relation:
sin2 β + cos2 β = 1 (by aforementioned relationship)
sin2 β = 1 - cos2 β
sin2 α sin2 β = sin2 α - sin2 α cos2 β (Multiply through by sin2 α)
but sin2 α = 1 - cos2 α
=> (1 - cos2 α)sin2 β = sin2 α - sin2 α cos2 β (Multiply through by sin2 α)
sin2 β - sin2 α + sin2 α cos2 β = sin2 β cos2 α as required.
Using this we can reduce our original equation to:
sin2 β cos2 α + 2 sin α cos β sin β cos α + sin2 β cos2 α = 2 sin β sin (α + β)cos θ
sin2 β cos2 α + sin α cos β sin β cos α = sin β sin (α + β)cos θ
cos α(sin β cos α + sin α cos β) = sin (α + β)cos θ (dividing through by sin β)
cos α sin (α + β) = sin (α + β)cos θ
cos α = cos θ
At last, our result! As you can see θ is purely a fuction of α, for all it's posturing in our equations β doesn't actually enter into it. That gives us Thm1. But look, it gets better! Since 0 ≤ α, θ ≤ 180 cos acts as a one to one function, so in fact α = θ, which is Thm 2.
Fortunately this is whole lot easier as we can use Thm 1 & 2. Because of Thm 1 it is adaquete to prove that α = β in the following diagram: . (We second right angle we have by the corollary of Thm 2.)
By "Angles on a line add up to 180" we have θ = 90 - α
By "Angles in a triangle add up to 180" we have:
β + θ = 90
β + 90 - α = 90
β = α as required.
Do give me feedback on this page, I'm interested to know how I can better present this. In particular on the algebra - I took the approach of giving lots of lines with explanations, because when reading maths in textbooks I often find it takes me ages to go from one line to another. Did I go too far, or not far enough?