Answer:
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.
