Two circles each of which passes through the centre of the other intersect points M and N.

Two circles each of which passes through the centre of the other intersect points M and N. A line from M intersects the circle at K and L as shown in the figure. If KL = 6 compute the area of ∆KLN.

I was attempting a Practice Paper and stumbled on this Question.

I think the tangent secant theorem must used somewhere which states that.

$$Tangent^2=Outer Secant×Whole Secant$$

But I don't know where to apply it so I am clueless.

I don't know where to start from.

Any help will be appreciated

Realize that $$\angle NKM= 120^o$$ and that $$\angle KLN= 60^o$$, which leaves $$\Delta KLN$$ an equilateral triangle. The answer is therefore $$9\sqrt{3}$$.
• I added another image as a hint. You already know that $\angle KLN= 60^o$, so according to the theorem I just added, $\angle KLN = 180^o - \angle KLN= 180^o- 60^o=120^o$. And the area of an equilateral triangle can be calculated if you know one side of the triangle. Commented Oct 17, 2019 at 11:27
• $\angle NOM=\angle NKM$ since these two angles are on the same intercepted arc. Commented Oct 17, 2019 at 11:31