Prove by vector method that the diagonals of a rhombus bisect each other. Also, show that they bisect each other at right angles
My Attempt:
Let us consider a rhombus $OACB$ where $\vec {OA}=\vec {a}$ and $\vec {OB}= \vec {b}$.
Then,
$$\vec {OC}=\vec {OA} + \vec {AC}$$
$$\vec {OC}=\vec {OA} + \vec {OB}$$
$$\vec {OC}=\vec {a} + \vec {b}$$
Similarly for $\vec {AB}$ we can write $\vec {AB}=\vec {b} – \vec {a}$.
Best Answer
We want to show that $$\frac12 \vec{OC} = \vec{OA}+\frac12\vec{AB}$$
The left hand side is $\frac12(\vec{a}+\vec{b})$.
The right hand side is $\vec{a}+\frac12(\vec{b}-\vec{a})=\frac12(\vec{a}+\vec{b})$.
Hence, we have shown that they bisect each other.