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Let us first draw the diagram of the question:

A is the centre of the circle and AM and AN are the radius of the circle.

$ \therefore $ AM = AN (1)

It is given that triangle ABC is an equilateral triangle. So, AB = BC = CA and $ \angle $ MBC = $ \angle $ NCB = $ \angle $ BAC = $ {60^0} $

Also, AM+MB = AN+NC

From equation 1, we get:

MB = NC.

Now, in $ \vartriangle MBC{\text{ and }}\vartriangle {\text{NBC}} $ , we have:

MB = NC

$ \angle $ MBC= $ \angle $ NCB = $ {60^0} $

And BC = BC

SO, by SAS rule of congruence, we can say that:

$ \vartriangle MBC \cong \vartriangle NBC $

And by c.p.c.t., we can say that CM = BN.

Therefore, we have proved that CM= BN.

1. SSS rule – All sides of one triangle are equal to all corresponding sides of another triangle.

2.SAS- Two sides and one angle formed between these two sides of one triangle is equal to that other triangle.

3. ASA- one side and two angles formed on this side of one triangle is equal to that of the other triangle.

4. RHS- This is applicable for the right triangle. Hypotenuse and one other side of one triangle is equal to that of another triangle.