Consider the following quadrilateral $ABCD$, with $E, F, G, H$ as the midpoint of $AD, DC, CB, BA$ respectively such that $\Delta ECH$ and $\Delta AGF$ are equilateral. Prove that $ABCD$ is a rhombus. Determine its angles.
I wanted to approach the first part of the problem using vector geometry. I started with the relations:
$$||\vec{HE}|| = ||\vec{EC}|| = ||\vec{CH}||$$
$$\implies ||\vec{H} - \vec{E}|| = ||\vec{C} - \vec{E}|| = ||\vec{H}-\vec{C}||\tag{1}$$
Similarly,
$$||\vec{A} - \vec{G}|| = ||\vec{G} - \vec{F}|| = ||\vec{F} - \vec{A}||\tag{2}$$
Now,
$$\vec{E} = \frac{\vec{A} + \vec{D}}{2}$$
$$\vec{H} = \frac{\vec{A} + \vec{B}}{2}$$
Putting these in $(1)$, we get:
$$\left|\left|\vec{C} - \frac{\vec{A} + \vec{D}}{2}\right|\right| = \left|\left|\vec{C} - \frac{\vec{A} + \vec{B}}{2}\right|\right|\tag3$$
Similarly,
$$\left|\left|\vec{A} - \frac{\vec{B} + \vec{C}}{2}\right|\right| = \left|\left|\vec{A} - \frac{\vec{D} + \vec{C}}{2}\right|\right|\tag4$$
Now, how do we go from there to saying:
$$\vec{A} + \vec{C} = \vec{D} + \vec{B}$$
And,
$$||\vec{D} - \vec{A}|| = ||\vec{A} - \vec{B}||$$
In general, how do I get past the absolute value operator in equations $(3)$ and $(4)$? I am a bit new to vector geometry and am completely lost on how to manipulate the absolute value operators.
Note that I know the solution to this using Euclidean geometry, so I want a solution using purely vector geometry here. At least for the first part of the question.
EDIT: I had an insight (I'm not sure if it's an insight yet or just a wild-goose chase): manipulating equation $(3)$ yields:
$$\left|\left|\vec{C} - \frac{\vec{A} + \vec{D}}{2}\right|\right|^2 = \left|\left|\vec{C} - \frac{\vec{A} + \vec{B}}{2}\right|\right|^2$$
$$\implies \left(\vec{C} - \frac{\vec{A} + \vec{D}}{2}\right)^2 = \left(\vec{C} - \frac{\vec{A} + \vec{B}}{2}\right)^2$$
Expanding and moving everything to the LHS and then factoring gives us:
$$\frac14(\vec{B} - \vec{D})(4\vec{C} - 2\vec{A} - \vec{D} - \vec{C}) = 0\tag5$$
Manipulating $(4)$ similarly gives us:
$$\frac14(\vec{B} - \vec{D})(4\vec{A} - 2\vec{A} - \vec{D} - \vec{B}) = 0\tag6$$
Dividing $(5)$ by $(6)$ and then cross multiplying, we get:
$$4\vec{C} - 2\vec{A} - \vec{D} - \vec{C} = 4\vec{A} - 2\vec{A} - \vec{D} - \vec{B}$$
$$\implies 3\vec{C} + \vec{B} = 4\vec{A}$$
Any ideas on how to go from here?