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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.

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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?

Gerard
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  • You can't divide $(5)$ by $(6)$ because $(5)=(6)=0$. Also, in terms of direction, $3\vec C+\vec B=4\vec A$ is plainly false simply by looking at your diagram – Tim Ratigan Nov 28 '13 at 12:45
  • Additionally, you there's a typo in $(5)$ and $(6)$. The last term should not be $\vec C$ but rather $\vec B$ in $(5)$, and in $(6)$ it should be $4\vec A-2\vec C$ – Tim Ratigan Nov 28 '13 at 12:48
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    Moreover, you're treating the dot product like scalar multiplication. It's fine to say that $\frac{1}{4}(\vec B-\vec D)\cdot (4\vec A-2\vec C-\vec D-\vec B)=0,$ but that just means that the two vectors being multiplied are orthogonal; you can't factor out $(\vec B-\vec D)$ by "dividing," rather, you only obtain that the two vectors that you multiply $\vec B-\vec D$ by are parallel. – Tim Ratigan Nov 28 '13 at 12:53

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