Solve the following problem....

Question

My problem is:

In a circle of radius $R$ is inscribed an equilateral triangle $ABC$. Through the point $C$ is drawn a line which intersects $AB$ in point $M$ and the circle, for the second time, in point $N$. Determine $CM\cdot CN$.

My idea is somehow to use sine theorem. So I wrote $CM$ as a function of $R$. But know I got stuck at finding $CN$ also as a function of $R$.

Thank you for help in advance!

Answer 1

On the picture below you can find two similar triangles, one containing another.

Triangles $CBM$ and $CNB$ are similar because $\angle CBM = \angle CNB$ and $\angle BCM$ is the common angle.

After you write down proportions of the sides of those triangles you will see that $CM\cdot CN$ doesn't really depend on the position of point $M$ on $AB$. It also should give you a simple formula for $CM\cdot CN$ as a function of $BC$.

$CM\cdot CN = a^2$ where $a$ is the length of the side of $\triangle ABC$.
It follows from $\dfrac{CM}{CB} = \dfrac{CB}{CN}$.

It seems that only $R$ is given to you so you have to also express your answer as a function of $R$.

Note that $BC = a = R\sqrt{3}.$

$CM\cdot CN = 3R^2$.