If P is a point and ABCD is a quadrilateral and vector(AP) + vector(PB) + vector(PD) = vector(PC), show that ABCD is a parallelogram.

Question

If P is a point and ABCD is a quadrilateral and \(\vec{AP}+\vec{PB}+\vec{PD}=\vec{PC}\), show that ABCD is a parallelogram.

Answer 1

Given a quadrilateral ABCD, P is a point outside the quadrilateral and

\(\vec{AP}+\vec{PB}+\vec{PD}=\vec{PC}\)

Consider ΔAPB and apply triangle law of vector, we get

\(\overset{\longrightarrow}{AP}+\overset{\longrightarrow}{PB}=\overset{\longrightarrow}{AB} \)...(ii)

And consider ΔDPC and apply triangle law of vector, we get

 \(\overset{\longrightarrow}{DP}+\overset{\longrightarrow}{PC}=\overset{\longrightarrow}{DC} \)...(iii)

Substitute the values from eqn(ii) an eqn(iii) in eqn(i), we get

\(\overset{\longrightarrow}{AB}=\overset{\longrightarrow}{DC} \)

Therefore, AB is parallel to DC and equal is magnitude.

Hence, ABCD is a parallelogram.

Hence proved

Comments

Firstly can't we prove it a parallelogram and then we can prove that statement