Problem: Five points O,A,B,C,D are taken in order on a straight line with distances OA=a,OB=b,OC=c, and OD=d. P is a point on the line between B and C and such that AP:PD=BP:PC. Then OP equals:
Answer Choices:
A. a−b+c−db2−bc
B. a−b+c−dac−bd
C. −a−b+c−dbd+ac
D. a+b+c+dbc+ad
E. a+b+c+dac−bd
Solution:
Since PDAP=PCBP,d−xx−a=c−xx−b
∴x(a−b+c−d)=ac−bd,x=a−b+c−dac−bd
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