Question

Let `M` be a point and `c` a circle. `M` doesn't belong to the `c`. Let `A` and `B` be points on `c`. `AM` has to be tangent to `c` and `BM` must cut `c` in a point `C`. How can I prove that `(d(A;M))^2 = d(B;M) * d(C;M)` ? (d is a distance function)

Answer 1

see explanation.

Explanation:

Proof of tangent-secant theorem.
See Fig 1.
Let `O` be the center of the circle.
Given `AM` is tangent to the circle, `=> angleMAO=90^@`
Let `angleMAC=y, => angleCAO=90-y`,
As `OA=OC=` radius, `=> angleACO=angleCAO=90-y`
`=> angleAOC=2y, => angleABC=y`
`=> angleMAC=angleABC`, a well-known property of tangents.

Now see Fig 2.
Let `angleAMC=x`,
as `angleMAC=angleABC=angleABM=y`,
`=> DeltaMAC and DeltaMBA` are similar.
`=> (CM)/(AM)=(AM)/(BM)`
`=> AM^2=BM*CM` --- (proved)

Footnotes : Tangent Secant theorem -- If a tangent line `AM` and a secant line `BCM` are drawn from a common point `M` outside the circle, then `AM^2=BM*CM`.