Fundamental Geometry Theorems and Proofs: Episode I.8

Today I am gonna go over another geometry theorem very closely related to the one solved in the last post, and I am going to prove it as well.


Here it is:
Let M be a point inside a circle C. The point M has a distance of a from the center of the circle C.
Let R be the radius of the circle. Then R > a.
Let any cord |AB| in the circle pass through M.
Prove that |MA|*|MB| = R^2 - a^2

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I now provide a proof using the Law of Cosines.


Let point O be the center of circle C.
Let angle OMA = θ

Then:


R^2 = a^2 + (|MA|^2) - 2a cos(θ)


R^2 = a^2 + (|MB|^2) - 2a cos(180 - θ)


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same as:

R^2 = a^2 + (|MA|^2) - 2a cos(θ)

R^2 = a^2 + (|MB|^2) + 2a cos(θ)

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then

R^2 - a^2 = (|MA|^2) - 2a cos(θ) = (|MB|^2) + 2a cos(θ)

hence:

(|MA| - |MB|)*(|MA| + |MB|) = 2a cos(θ)*(|MA| + |MB|)

simplifying some terms we get:

(|MA| - |MB|) = 2a cos(θ)

and then

|MA| = |MB| + 2a cos(θ)

multiply both sides by |MA| and:

|MA|^2 = |MA|*|MB| + 2a |MA| cos(θ)

So:

|MA|*|MB| = |MA|^2 - 2a |MA| cos(θ)

therefore:

|MA|*|MB| = R^2 - a^2

QED time!!!