only answer if you know and can give an explanation, will report for non-sense answers

Below

Step-by-step explanation:

For a given shape to be a rhombus, it should satisfy these conditions:

● The diagonals should intercept each others in the midpoint.

● The diagonals should be perpendicular.

● The sides should have the same length.

We will prove the conditions one by one.

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Let's prove that the diagonals are perpendicular:

To do that we will write express them as vectors

The two vectors are EG and DF.

The coordinates of the four points are:

● E(0,2c)

● G (0,0)

● F (a+b, c)

● D (-a-b, c)

Now the coordinates of the vectors:

● EG (0-0,0-2c) => EG(0,-2c)

● DF ( a+b-(-a-b),c-c) => DF (2a+2b,0)

For the diagonals to be perpendicular the scalar product of EG and DF should be null.

● EG.DF = 0*(2a+2b)+(-2c)*0 = 0

So the diagonals are perpendicular.

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Let's prove that the diagonals intercept each others at the midpoints.

The diagonals EG and DF should have the same midpoint.

● The midpoint of EG:

We can figure it out without calculations. Since G is located at (0,0) and E at (0,2c) then the distance between E and G is 2c.

Then the midpoint is located at (0,c)

● The midpoint of DF:

We will use the midpoint formula.

The coordinates of the two points are:

● F (a+b,c)

● D(-a-b,c)

Let M be the midpoint of DF

●M( (a+b-a-b,c+c)

● M (0,2c)

So EG and DF have the same midpoint.

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There is no need to prove the last condition, since the two above guarante it.

But we can prove it using the pythagorian theorem.

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