ABC is a triangle and P any point on BC. if ` vec P Q` is the sum of ` vec A P` + ` vec P B` +` vec P C` , show that ABPQ is a parallelogram and Q , therefore , is a fixed point.

Explanation

Here `" "vec(PQ)=vec(AP)+vec(PB)+vec(PC)`
`" "vec(PQ)-vec(PC)=Avec(AP)+vec(PB)`
`" "vec(PQ)+vec(CP)=Avec(AP)+vec(PB)`
`" "vec(CQ)=vec(AB)rArrCQ=AB and CQ "||"AB`
Therefore, ABQC is a parallelogram.
But A, B and C are given to be fixed points and ABQC is a parallelogram. Therefore, Q is a fixed point.