By using vector method prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the square of its sides.

please answer asap!!!

Question

By using vector method prove that the sum of the squares of the
diagonals of a parallelogram is equal to the sum of the square of
its sides
please answer asap!!!

Gursheen kaur. · Accepted Answer

Let ABCD be a parallelogram Let A be the origin and the position vectors of B and D be $\vec{a} a n d \vec{b}$ One diagonal ,AC= $\vec{d_{1}} = \vec{a} + \vec{b}$ and the other diagonal ,BD= $\vec{d_{2}} = \vec{a} - \vec{b}$ As, the square of the magnitude of a vector is equal to its scalar product by itself ${\vec{d_{1}}}^{2} = \vec{d_{1}} \vec{d_{1}} = (\vec{a} + \vec{b}) (\vec{a} + \vec{b}) = {\vec{a}}^{2} + {\vec{b}}^{2} + 2 \vec{a} \vec{b} {\vec{d_{2}}}^{2} = \vec{d_{2}} \vec{d_{2}} = (\vec{a} - \vec{b}) (\vec{a} - \vec{b}) = {\vec{a}}^{2} + {\vec{b}}^{2} - 2 \vec{a} \vec{b} N o w, \Rightarrow {\vec{d_{1}}}^{2} + {\vec{d_{2}}}^{2} = {\vec{a}}^{2} + {\vec{b}}^{2} + 2 \vec{a} \vec{b} + {\vec{a}}^{2} + {\vec{b}}^{2} - 2 \vec{a} \vec{b} \Rightarrow {\vec{d_{1}}}^{2} + {\vec{d_{2}}}^{2} = {\vec{a}}^{2} + {\vec{b}}^{2} + {\vec{a}}^{2} + {\vec{b}}^{2} H e n c e, t h e r e s u l t$

By using vector method prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the square of its sides. please answer asap!!!

By using vector method prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the square of its sides.

please answer asap!!!