show that the quadrilateral is a parallelogram if and only if its diagonals bisect each other - Maths - Vector Algebra

Question

show that the quadrilateral is a parallelogram if and only if
its diagonals bisect each other

Chetna Khanna · Accepted Answer

Let, ABCD be a parallelogram To prove: Diagonals of quadrilateral ABCD bisect each other Let the position vector of A, B, C and D be $\vec{a}, \vec{b}, \vec{c} and \vec{d}$ respectively Since ABCD is a parallelogram $∴ \vec{AB} = \vec{DC} \Rightarrow \vec{b} - \vec{a} = \vec{c} - \vec{d} \Rightarrow \vec{b} + \vec{d} = \vec{c} + \vec{a} \Rightarrow \frac{1}{2} (\vec{b} + \vec{d}) = \frac{1}{2} (\vec{c} + \vec{a})$ â‡’ Position vector of the mid point of BD = Position vector of the mid point of CA

Thus, the point which bisects BD also bisects CA Hence, diagonals of parallelogram ABCD bisect each other Conversely, Let ABCD be a quadrilateral whose diagonals bisect each other To prove: ABD is a parallelogram Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of the vertices A, B, C, D respectively Since diagonals AC and BD bisect each other âˆ´ Position vector of the mid point of AC = Position vector of the mid point of BD $\Rightarrow \frac{1}{2} (\vec{a} + \vec{c}) = \frac{1}{2} (\vec{b} + \vec{d}) \Rightarrow \vec{a} + \vec{c} = \vec{b} + \vec{d} (1) \Rightarrow \vec{a} - \vec{b} = \vec{d} - \vec{c} \Rightarrow \vec{BA} = \vec{CD}$ Again from (1), $\vec{a} - \vec{d} = \vec{b} - \vec{c} \Rightarrow \vec{DA} = \vec{CB}$ Hence, ABCD is a parallelogram

show that the quadrilateral is a parallelogram if and only if its diagonals bisect each other.