A difficult geometry problem for smart mathematicians!?

Question

Suppose M is the center of a circle and A and B two points on the circle circumference. (AB is not a diameter)

The tangent lines on the circle from A and B intersect in C.

CM intersects with the circle in D.

The tangent line of circle from D, intersects with AC and BC in E and F respectively.
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Please show that the area of ADBM is the geometric mean of the areas of ABM and ACBM.

Please see this image:
http://m1.freeshare.us/156fs850081.jpg

Answer 1

Mark intersection of AB and MC with G. Then from the similarity of MAC and MAG follows:
MA/MC = MG/MA, and since MD = MA follows:
MD/MC = MG/MD. If you multiply all by AG you get:
(MD*AG)/(MC*AG) = (MG*AG)/(MD*AG).
MD*AG = 2*AMD = ADBM
MC*AG = 2*AMC = ACBM
MG*AG = 2*AMG = ABM
Therefore: ABM * ACBM = ADBM^2.

Answer 2

I assume that the geometric mean of the two areas is like the mean of a set of numbers.

In this case maybe ADBM = ABM + ACBM / 2

Ok, Im no mathematician, but I did a similar problem to this a while back. You need to get the formulas

Area of a sector: 0.5 r^2*X (where X is the angle)
Area of a triangle: 0.5 (base)(height)

and do subsitution. I'll have a look at it now and see how I get on. Ouch tough one.

ABM = pi*r^2 - 0.5 r^2*X

Answer 3

Proof
Draw segment AB intersecting DM at G.
By symmetry, segment AB ┴ segment DB and GB is the common altitude of ∆ BGM and ∆ CBM
∆ BGM ~ ∆ CBM (AA~ theorem)
MG/DM = DM/CM, because DM = MB = radius
(DM)^2 = (MG)(CM)
Multiply both sides by GB^2
(DM*GB) = (MG*GB)(CM*GB)
Therefore, we proved
(the area of ADBM)^2 = (the area of ABM)(the area of ACBM
)