Answer to Question #157074 in Geometry for Kelvin Sowah

Question

ABC is a triangle and P is any point in BC. If PQ is the resultant of AP, PB, PC, show that ABQC is a parallelogram, and Q is therefore a fixed point

Answer 1

1) By adding all of the vectors we get the point C` and the vector AC, which, when translated, results in the vector PQ at the last drawing.

2) AP = C`Q (the rule of translation), BC` = PC (when added vectors BP was translated into CC` vector). C`Q is parallel to AP so that the C` angle equals to the P angle. Thus, the C`BQ triangle equals the PCA triangle. Consequently, BQ = AC, similarly, AC = BQ, ABQC is a parallelogram.