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)

1 Answer
Sep 3, 2017

see explanation.

Explanation:

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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#.