I've been bugged by a problem my little sister's maths teacher gave her today.
The question is : Prove that the diagonals of a rhombus are perpendicular
This exercice is in the vector and scalar product section, so I guess the teacher is expecting it to be solved using that subject.
What I've worked out so far is that if we take both diagonals as vectors, they are perpendicular if their scalar product is equal to zero.
Which leads to the affirmation
$$
\left\|CA\right\| * \left\|DB\right\| * \cos(\theta) = 0
$$
Since $\left\|CA\right\|$ and $\left\|DB\right\|$ are different from 0 it means that $\cos(\theta)$ equals zero and then $\theta$ is equal to 90°.
I couldn't go any further. How can I prove that $\theta$ is 90° given only the fact that it is an angle generated by a rhombus' diagonals ?
Best Answer
If you let the vector AB be $x$ and AD be $y$, and point A be the origin, then we know AC is $x+y$ and DB is $x-y$. Now the dot product between AB and AD is $$(x+y)\cdot (x-y)$$ $$=x\cdot x-x\cdot y+y\cdot x-y\cdot y$$ $$=|x|^2-|y|^2=0$$ since dot product is symmetric and $|AB|=|AD|$. Thus AB and AD are perpendicular.