Prove by vector method that the diagonals of a rhombus bisect each other. Also, show that they bisect each other at right angles.

Question

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 $\vec {AB}=\vec {b} - \vec {a}$.

Answer 1

To show that the diagonals cross at right angles you need to show that

$\vec{OC}\cdot\vec{AB} = (\vec a + \vec b) \cdot (\vec b- \vec a)=0$

You will need to use the fact that $|a|=|b|$, because this property is true for a rhomubs but not for a general parallelogram.

To show that the diagonals bisect one another you need to express $\vec b$ as a linear combination of $\vec a+ \vec b$ and $\vec b - \vec a$. In other words find constants $\alpha$ and $\beta$ such that

$\vec b = \alpha(\vec a + \vec b) +\beta (\vec b- \vec a)$

Note that this property is true for a general parallelogram, so it does not depend on the fact that $|a|=|b|$ for a rhombus.

Answer 2

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.