Properties of Parallelograms Practice with Our Examples

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
$$\begin{gathered} \mathbf{\text{(A)}}\text{a square}\mathbf{\text{(B)}}\text{a rectangle}\mathbf{\text{(C)}}\text{a parallelogram with unequal sides} \\ \mathbf{\text{(D)}}\text{a rhombus}\mathbf{\text{(E)}}\text{a quadrilateral with no special properties} \end{gathered}$$

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 $\stackrel{\to <!--\; \to \; -->}{OA}=\stackrel{\to <!--\; \to \; -->}{a}$ and $\stackrel{\to <!--\; \to \; -->}{OB}=\stackrel{\to <!--\; \to \; -->}{b}$. Then,
$\stackrel{\to <!--\; \to \; -->}{OC}=\stackrel{\to <!--\; \to \; -->}{OA}+\stackrel{\to <!--\; \to \; -->}{AC}$
$\stackrel{\to <!--\; \to \; -->}{OC}=\stackrel{\to <!--\; \to \; -->}{OA}+\stackrel{\to <!--\; \to \; -->}{OB}$
$\stackrel{\to <!--\; \to \; -->}{OC}=\stackrel{\to <!--\; \to \; -->}{a}+\stackrel{\to <!--\; \to \; -->}{b}$
Similarly for $\stackrel{\to <!--\; \to \; -->}{AB}$ we can write $\stackrel{\to <!--\; \to \; -->}{AB}=\stackrel{\to <!--\; \to \; -->}{b}-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{a}$.

"The sum of the squares of the diagonals is equal to the sum of the squares of the four sides of a parallelogram."
I find this property very useful while solving different problems on Quadrilaterals & Polygon,so I am very inquisitive about a intuitive proof of this property.

Given a lattice $\mathrm{\Gamma <!--\; \Gamma \; -->}\subset <!--\; \subset \; -->\mathbb{C}$, a Theta function $\mathrm{\vartheta <!--\; \vartheta \; -->}:\mathbb{C}\to <!--\; \to \; -->\mathbb{C}$ is a holomorphic function with the following property:
$\mathrm{\vartheta <!--\; \vartheta \; -->}(z+\mathrm{\gamma <!--\; \gamma \; -->})=e^{2i\mathrm{\pi <!--\; \pi \; -->}{a}_{\mathrm{\gamma <!--\; \gamma \; -->}}z+b_{\mathrm{\gamma <!--\; \gamma \; -->}}}\mathrm{\vartheta <!--\; \vartheta \; -->}(z)$
for every $\mathrm{\gamma <!--\; \gamma \; -->}\in <!--\; \in \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}$, and $a_{\mathrm{\gamma <!--\; \gamma \; -->}},b_{\mathrm{\gamma <!--\; \gamma \; -->}}\in <!--\; \in \; -->\mathbb{C}$.

Exercise: A Theta function never vanishes iff $\mathrm{\vartheta <!--\; \vartheta \; -->}(z)=e^{p(z)}$ with $p(z)$ a polynomial of degree at most 2.

Hint: The "only if" part is trivial. The hint is: show that $\log <!--\; \; -->(\mathrm{\vartheta <!--\; \vartheta \; -->}(z))=O(1+|z{|}^2)$. I tried to apply log on both sides, or derive one and two times, or everything I could have thought of. I don't get where the square comes from.

If you draw a parallelogram its diagonals will form four triangles of which at least the opposite pairs will be congruent, which is to say, each triangle is a reflection of its opposite one, why is this? What property of parallelograms am I missing here? because if I draw a trapezium its diagonals will obviously not form congruent triangles.

Consider the following wave equation:
$\begin{array}{rl}{u}_{tt}-<!--\; -\; -->u_{xx}& =0,0<t<tx,k>1\\ u{|}_{t=0}& ={\mathrm{\varphi <!--\; \varphi \; -->}}_0(x),x\ge <!--\; \ge \; -->0\\ u_t{|}_{t=0}& ={\mathrm{\varphi <!--\; \varphi \; -->}}_1(x),x\ge <!--\; \ge \; -->0\\ u{|}_{t=kx}& =\mathrm{\psi <!--\; \psi \; -->}(x)\end{array}$
In which ${\mathrm{\varphi <!--\; \varphi \; -->}}_0(0)=\mathrm{\psi <!--\; \psi \; -->}(0)$.

The problem is that on part of the sector where $t>x$ d'Alembert's formula isn't applicable. It seems we will have to do some sort of extension or reflection, but how to start?

Prove if $|z|<1$ and $|w|<1$, then $|1-<!--\; -\; -->z{w}^{\ast <!--\; \ast \; -->}|\ne <!--\; \ne \; -->0$ and $|\frac{z-<!--\; -\; -->w}{1-<!--\; -\; -->z{w}^{\ast <!--\; \ast \; -->}}|<1$
Given that $|1-<!--\; -\; -->z{w}^{\ast <!--\; \ast \; -->}{|}^2-<!--\; -\; -->|z-<!--\; -\; -->w{|}^2=(1-<!--\; -\; -->|z{|}^2)(1-<!--\; -\; -->|w{|}^2)$
I think the first part can be proven by saying $|1-<!--\; -\; -->z{w}^{\ast <!--\; \ast \; -->}|=0$ if and only if $z{w}^{\ast <!--\; \ast \; -->}$ = 1.

And given the conditions that cannot be true. However I don't know if this part is right.

If a line L separates a parallelogram into two regions of equal areas, then L contains the point of intersection of the diagonals of the parallelogram.
The figure shows a line L horizontally through the sides of the parallelogram.
This creates two trapezoids and I can intuitively show that if the bases of the trapezoids are not congruent then the areas can not be equal.
I just can not currently see how to relate the intersection with the diagonals.
Any help will be appreciated.

I've been away from math far too long, and now my memory plays tricks on me when I try to recall simple facts.

I know that $det(v_1,\dots <!--\; \dots \; -->,v_n)$ is the oriented volume of the simplex determined by the origin and the vectors $v_1,\dots <!--\; \dots \; -->,v_n$ (up to a constant factor depending on the dimension $n$).

But I fell into utter confusion when I tried to make up why, especially when I tried to prove that this formula works for a shifted simplex, too.

What I'm sure is that det is an alternating bi-linear form that fits naturally with an oriented volume function.

What I've found on the net is dependent on what definition is used, and prone to circular reasoning.

Explicit question: how and why is the oriented volume of a simplex related to the determinant of its vectors?

What I've tried: proved it for $n=1,n=2,n=3$, and I've seen that this is not the way to go.

1) I know that all inner product space is also a normed space with the norm induce by the scalar product, but is the reciprocal true ? I mean, is all normed space also a inner product space ?
2) I know that all normed space is a metric space with the metric induced by the norm. Is the reciprocal true ? I mean, is all metric space also a normed space ?