Proofs for the Circle Theorems 
When I did GCSE maths 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 unaccountably failed to disturb me at the time. Some time later I came across them again and wanted proofs. Unable to find proofs on the internet, I discovered proofs for myself.
I shall need some terminology in order to state the theorems, so I suppose I had better define those terms.
It is also worth noting that I shall be working in degrees for these proofs. Radians are better if you're doing calculus, but we're not.
Let's start by giving the angles different names (otherwise we'd just be assuming we were right, instead of proving it). Then our task is to prove that θ = α (I've made the angle at the middle 2α instead of α just to make the numbers a bit easier). We can't work anything out yet, but if we split θ in 2 then we can.  
Now everything's a triangle, and more than that all 3 triangles have two sides that are the radius of the circle, making them the same length. That means all our triangles are isosceles triangles. The two angles at the bottom of an isosceles triangle are equal (if you draw a line down the middle of an isosceles triangle then it's a line of symmetry) so we can label two more angles.  
The angles in a triangle add up to 180° so we can now label the angles at the centre.  
The angles at the centre must add up to 360° since that's how many degrees it takes to go all the way round. Therefore 2α + 180°  2β + 180°  2γ = 360° cancelling the 360° on each side 2α  2β  2γ = 0 divide by 2 α = β + γ but β + γ = θ, so α = θ  the thing we were trying to prove to start with. 
This looks great, but unfortunately we cheated  we assumed that the centre of the circle was inside the triangle made by the angle at the circumfrence and the chord. What if it isn't?
Well we can still draw in the radius.  
That gives us a large isosceles triangle on the left. Let's call the angle at the base β. Since it's an isosceles we know the other angle at the base is also β, and by all angles in a triangle add up to 180° we know the angle at the centre is 180°  2β  
Now we've got another isosceles triangle at the top if we take together the 180°  2 β and the 2α. What is the angles at the base of this triangle? Well they have to be the same as it's an isosceles triangle, and they have to add up to 180° 180°  2 β + 2α + 2base = 180° base = β  α 

Look at the angle on the left. We have it as a whole (β) and split in two (βα and θ) therefore: β  α + θ = β θ = α 
Theorem 1 holds whether the centre is inside our triangle or not!
Fortunately this is whole lot easier as we can use Theorem 1. Take any two angles θ and φ subtended by the same chord. That chord subtends one angle at the centre (2α).
By theorem 1 θ = α and φ = α => θ = φ
Fortunately this is whole lot easier as we can use Theorem 1 & 2. Because of Theorem 1 it is adaquete to prove that α = β in the following diagram: . (We second right angle we have by the corollary of Theorem 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.
This page originally had much more complicated proofs of the same theorems. It seemed a shame not to preserve these for the amusement of posterity, so I present to you Unneccessarily Complex Circle Theorem Proofs
©Neil Roques 2017
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