If $A,B,C,D$ are the vertices of a rhombus and $M$ is the point of intersection of the diagonals such that $2AC=BD$, find the vertex $A$ if $D$ is $1+i$ and $M$ is $2-i$.
Attempt:
$$MA\perp MD$$ $$MA\times MD=\pm i\vert MA\vert\cdot\vert MD\vert$$ $MD=\sqrt{5}$ and $MA=\frac{\sqrt{5}}{2}$
Let $A$ be $a+bi$
After taking the cross product, $4a+2b-11=0$ or $4a+2b-1=0$
How should I proceed?
Translating the problem with the vector $i-2$ we obtain $M=0$ and $D=2i-1$. Because $B$ is symmetric to $D$ (respect to $M$), then $B=1-2i$. Now, the symmetric of $B$ respect to axis x has the same modulo and is $1+2i$, because $2AC=BD$ we obtain that $C=\frac{1+2i}{2}$ and $A$ is the symmetric of $C$, $ A=\frac{-1-2i}{2}$.
Finally, translating to original position (adding 2-i) we obtain the solution.
Comment: you can change A to C and is the same