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Prove that the diagonals of a parallelogram bisect each other and

that the diagonals of a rhombus are orthogonal.

Complex numbers has some similarities with vectors, if along one of the vector's methods:

$$\begin{align*}\frac{z_1+z_2}{2}+\frac{z_1-z_2}{2}=z_1\end{align*}$$

。。。

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one method I'm trying:(Is it feasible?)

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$z_1,z_2,z_3,z_4$ is four vertexes of the parallelogram denoted in succession.

$z_1z_2\text{//}z_3z_4$ or $\text{//} z_4z_3$

the parameter of the two lines is $z=a+b t=a'+b't=a'+b c t$, where $a,b$ are complex numbers, and $t$ is taken over all real numbers, $c$ is a real number.

then solve for the line $z_1z_3$, and $z_2z_4$, and then solve for the intersection denoted by $p$.

(how to solve for $p$? Should I Set $z_1=x_1+y_1i$... and so forth?).

to prove:

$$\begin{align*}\left|p-z_2\right|=\left|p-z_4\right|\\&\text{and}\\\left|p-z_1\right|=\left|p-z_3\right|\end{align*}$$

I'm leaning complex numbers, and also you can add your interesting answers.

I'm new here, please be kind.

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4 Answers

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The conventions to using complex numbers to analytically analyze plane geometry problems are as follows.

First, pick a suitable origin -- in this case probably one of the vertices of your quadrilateral, say $A$ in $ABCD$. Then define complex numbers to represent all the "wiggly bits" of your diagram -- here you will want to normalize one edge of your parallelogram to be $1$, i.e. WLOG let $B = 1 \in \mathbb C$, and let $D = z \in \mathbb C$.

Now the next step is to compute whatever it is you want to find in 2 ways and verify that the complex expressions match. Here we want to compute the midpoints of $AC$ and $BD$ and verify that they are the same.

$C$ is located at $1+z$, so to find $AC$ we simply use the midpoint formula for complex numbers:
$z_M = \dfrac{z_P + z_Q}2$ iff $M$ is the midpoint of $PQ$.

There are plenty of these kinds of formulas to verify collinearity, concurrency, perpendicularity/parallelity of lines, cyclicity of quads, etc. They can be rederived or looked up.

For instance, in the second part, we specify that $ABCD$ is additionally a rhombus; this is equivalent to setting $|z| = 1$ or $z \bar z = 1$. To test that the diagonals are orthogonal, you can use the rule
$ i \dfrac{ z_P - z_Q }{z_R - z_S} \in \mathbb R$ iff $PQ \perp RS$, i.e.
$ \dfrac{z_P - z_Q }{ z_R - z_S} = -\overline { \left ( \dfrac{z_P- z_Q }{ z_R - z_S} \right )}. $

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Let E(e) be the point of bisection of the diagonals Also since ABCD is a parallelogram Vector AB =Vector DC Therefore b-a=c-d b+d=c+a b+d/2= c+a/2=e

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Assume that the vertices of the parallelogram are labelled as $z_1$, $z_2$, $z_3$, and $z_4$ counterclockwise with $z_1$ at the bottom left corner. Then, since ${z_2}-{z_1}={z_3}-{z_4},$ it follows that ${z_1}-{z_4}={z_2}-{z_3}.$ Thus, $${\frac{{z_1}+{z_3}}{2}}={\frac{{z_2}+{z_4}}{2}}.$$Thus the midpoints of the diagonals of the parallelogram coincide, which implies that the diagonals must bisect each other. If the parallelogram is a rhombus, and the common midpoint of both diagonals is $m$, then the triangles ${\Delta}{z_1}{z_2}m$ and ${\Delta}{z_2}{z_3}m$ are congruent since ${z_2}-{z_1}={z_3}-{z_2},$ the side $m-{z_2}$ is common to both triangles and (as the diagonal ${z_1}+{z_3}$ is bisected), $|m-{z_1}|=|{z_3}-m|.$ We now observe that ${\angle}{z_1}m{z_2} = {\angle}{z_3}m{z_2}$ being opposite to the adjacent sides ${z_2}-{z_1}$ and ${z_3}-{z_2}$ of the given rhombus which are of equal length. Since the sum of both angles is $\pi$ radians, each must measure ${\pi}/2$ radians.

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Assume that The parallelogram is formed by two complex numbers $a$ and $b$, not on the same ray through the origin, so that its vertices are $0$, $a$, $a+b$, and $b$ read counterclockwise starting at $0$. Then we seek real numbers $\lambda$, and $\mu$ both in $[0,1]$ such that $${\lambda}(a+b) =a+{\mu}(b-a).$$ This is easily seen to be the point at which the diagonals meet. The above displayed equation can be written as $$(1-({\lambda}+{\mu}))a+({\mu}-{\lambda})b=0.$$ Since the two vectors $a$ and $b$ are linearly independent, the only possibility is that $$(1-({\lambda}+{\mu}))=0$$ which implies that ${\lambda}+{\mu}=1$, and ${\mu}-{\lambda}=0$ forcing $${\lambda}={\mu}={\frac {1}{2}}.$$

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