Question 618985: Given: triangle ABC with vertices A(-6,-2), B(2,8), and C(6,-2). Line segment AB has midpoint E, and line segment AC has a midpoint F. Prove: ADEF is a parallelogram and ADEF is NOT a rhombus.
Answer by solver91311(24713)   (Show Source):
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Point D has to be at (-2,3) or (-4,-7) if indeed ADEF is a parallelogram, and it makes no difference which, the quadrilateral is not a rhombus in either case.
Use the mid-point formulas to determine the coordinates of point E and F.
 and
I'll leave it to you to verify E(-2,3) and F(0,-2)
Then using the properties of a parallelogram and the fact that parallel lines have equal slopes, we can determine that point D is either (-2,3) or (-4,-7).
If ADEF were a rhombus, then either AE = EF or AF = EF (depending on where D is located)
Use the distance formula to disprove each assertion. ADEF not a rhombus, Q.E.D
Distance formula:
^2\ +\ (y_1\ -\ y_2)^2})
John
My calculator said it, I believe it, that settles it
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