GCSEs are over and the holiday is just kicking off. Time to return to WordPress with some trivial and not-so-trivial lemmata!

Here is what I discovered while doing some maths. I will give two proofs of this result – one using angles, and another using lengths (invoking some more advanced technology). Which do you prefer?

Circle A is internally tangent to larger circle B at the point Y. Points X and Z lie on the circle B such that the line XZ is tangent to circle A at the point M. **Then MY bisects angle XYZ.
**

**Let O**

Proof 1

Proof 1

_{A }be the centre of A and let O

_{B}

_{ }be the centre of B.

Observe that, since Y is the point of tangency between A and B, O

_{B}, O

_{A}

_{ }and Y are collinear.

Let ∠MYO

_{A }= α and let ∠O

_{B}YZ = β. We want to show that ∠XYM = α + β.

O

_{A}M = O

_{A}Y (equal radii) so triangle O

_{A}MY is isosceles; therefore, ∠O

_{A}MY = α

Tangent and radius are perpendicular so ∠XMY = 90 – α

Also, O

_{B}Y = O

_{B}Z (equal radii) so triangle O

_{B}YZ is isosceles; therefore, ∠O

_{B}ZY = β

Since angles in a triangle sum to 180, ∠ZO

_{B}Y = 180 – 2β

∠MXY = ½∠ZO

_{B}Y = 90 – β (angle at centre is twice angle at circumference)

By angles in a triangle sum to 180, ∠XYM = 180 – ∠XMY – ∠MXY = α + β as required

Hence MY bisects angle XYZ.

**Proof 2
**Let XY and ZY cut the circle A at X’ and Z’ respectively, as shown:

Since the circles A and B touch at Y, we can consider Y as the centre of the enlargement that transforms circle A into circle B. Let the enlargement have scale factor K; then

Or

And

So

which rearranges to

Also, by power of a point/tangent-secant theorem/whatever you call it:

And

Dividing the first equation by the second,

Square rooting:

Then, by the converse of the angle bisector theorem, MX bisects angle XYZ.

We can see that X’Z’ looks parallel to XZ. This can also be proved in two ways – firstly, by angles, using the above result in conjunction with alternate segment theorem – or, much more simply, by using a homothety/enlargement argument.