proving that an angle in a triangle is 60 if and only if a certain condition satisfied I have the following problem with me:
" Let ABC be an acute angled triangle with angle bisectors BL and CM, with L lying on AC and M lying on AB Prove that angle A = 60 if and only if there exists a point K on BC(K is neither B nor C) such that triangle KLM is equilateral"
I am amazed as to how the angle of a triangle can be determined by such a property. Tried angle chasing but didn't work. Any ideas?
 A: Part One
Let us first prove that if $\angle A=\alpha=60^\circ$, you can construct an equilateral triangle $MLK$:

$$\angle LSM=\angle CSB=180^\circ-\frac\beta2-\frac\gamma2=180^\circ-\frac{180^\circ-\alpha}{2}=90^\circ+\frac\alpha2=120^\circ$$
This means that quadrilateral $LSMA$ is cyclic (sum of opposite angles $\angle LSM$ and $\angle LAM$ is $180^\circ$).
This implies: 
$$\angle SLM=\angle SML=\angle LAS=\angle MAS=30^\circ\\LS=LM$$
WLOG, suppose that $CA\le CB$. This means that $CL<CA$. Find point K symmetric with respect to bisector $CM$. That point lies between points BC (because $CK=CL<CA\le CB$).
Obviously $ML=MK$ and triangle $MLK$ is isosceles with $\angle MLK=\angle MKL$. On the other side $LK\bot MC$ and $\angle LMC=30^\circ$ so $\angle MLK=\angle MKL=60^\circ$. So the triangle $MLK$ is equilateral.
PART TWO
Let us prove the opposite statement. We'll start with equilateral triangle $KLM$ and prove that $\angle A=60^\circ$. 

The locus of points B such that $\angle LBM=\frac\beta2$ is the blue arc (a part of a cirlce with some radius $R$). The locus of points B such that $\angle LBK=\frac\beta2$ is the green arc (a part of a different cirlce). Because $\angle LBM = \angle LBK$ and $LK=LM$, these two arcs (circles) have the same radius and are symmetric with respect to common chord $LB$. 
This means that $LB\bot MK$. In a similar way you can show that $MC\bot LK$. So angles $\angle CSB$ and $\angle LKM$ have perpendicular legs with one of them being acute and the other one obtuse. So it means that:
$$\angle CSB=180^\circ-\angle LKM=120^\circ\tag{1}$$
But in part one we already proved that:
$$\angle CSB=90^\circ+\frac\alpha2\tag{2}$$
From (1) and (2) you get that $\alpha=60^\circ$.
