In a triangle ABC, points D and E are on segments BC and AC such that BD = 3DC and AE = 4EC. Point P is on line ED such that D is the midpoint of segment EP. Lines AP and BC intersect at point S. Find the ratio BS/SD (RMO 2013, Mumbai Region). Please give solution with figure.
Let $F$ denote the midpoint of the segment $A E .$ Then it follows that $D F$ is parallel to $A P .$ Therefore, in triangle $A S C$ we have $C D / S D=C F / F A=3 / 2 .$ But $D C=B D / 3=$ $(B S+S D) / 3 .$ Therefore $B S / S D=7 / 2$
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