ABC is an acute angled triangle. P,Q are the points on AB and AC respectively such that area of ΔAPC = area of ΔAQB. A line is drawn through B parallel to AC and meets the line trough Q parallel to AB at *. QS cuts BC at R. Prove that RS = AP.
As ΔPQC ΔQPB have same area & same base, so they must lie between same parallel PQ || BC
As AQ || BS and AB || SQ
AP = RS (CPCT)
