Different points $A, B, C, D$ lie on a circle with a center at the point $O$ at such way that $\angle AOB$ $= \angle BOC =$ $\angle COD =$ $60^o$. Point $P$ lies on the shorter arc $BC$ of this circle. Points $K, L, M$ are projections of $P$ on lines $AO, BO, CO$ respectively . Show that (a) the triangle $KLM$ is equilateral, (b) the area of triangle $KLM$ does not depend on the choice of the position of point $P$ on the shorter arc $BC$