Question

In any $△ABC$, a point $P$ is on the side $BC$. If $\overrightarrow{PQ}$ is the resultant of the vector $\overrightarrow{AP},\overrightarrow{PB}$ and $\overrightarrow{PC}$ then prove that $ABQC$ is a paralleogram and hence $Q$ is a fixed point.

Answer 1

Given that:-

$\overrightarrow{PQ}=\overrightarrow{PQ}=\overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PC}$

To prove:-

$ABCQ$ is a parallelogram.

Proof:-

$\overrightarrow{PQ}=\overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PC}$

$\overrightarrow{PQ}-\overrightarrow{PC}=\overrightarrow{AP}+\overrightarrow{PB}$

$\Rightarrow \overrightarrow{PQ}+\overrightarrow{CP}=\overrightarrow{AP}+\overrightarrow{PB}$

$\Rightarrow \overrightarrow{CQ}=\overrightarrow{AB}$

$\therefore AB=CQ\&AB\parallel CQ$

Hence the quadrilateral $ABCQ$ is a parallelogram.

Since $A,B$ and $C$ are given points of the $△ABC$ are fixed so the point $Q$ is also a fixed point.

Hence proved.