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 OA be the centre of A and let OB be the centre of B.
Observe that, since Y is the point of tangency between A and B, OB, OA and Y are collinear.
Let ∠MYOA = α and let ∠OBYZ = β. We want to show that ∠XYM = α + β.
OAM = OAY (equal radii) so triangle OAMY is isosceles; therefore, ∠OAMY = α
Tangent and radius are perpendicular so ∠XMY = 90 – α
Also, OBY = OBZ (equal radii) so triangle OBYZ is isosceles; therefore, ∠OBZY = β
Since angles in a triangle sum to 180, ∠ZOBY = 180 – 2β
∠MXY = ½∠ZOBY = 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.
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
which rearranges to
Also, by power of a point/tangent-secant theorem/whatever you call it:
Dividing the first equation by the second,
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.