Question 554048: This is from the geometry regents August 2011, number 38.
Given: Triangle ABC with vertices A(-6,-2), B(2,8), and C(6,-2)
Line AB has midpoint D, Line BC has midpoint E, and Line AC has midpoint F.
Prove: ADEF is a parallelogram.
ADEF is not a rhombus.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Was this an open-ended question?
There is more than one way to solve the problem.
The coordinates of the vertices of the triangle are irrelevant to proving ADEF is a parallelogram. Without giving coordinates, they could have told you that triangle ABC is not isosceles, and let you prove ADEF is not a rhombus using pure geometry.
The fact that they give you coordinates at all means that either they want to give alternate ways to solve it for those not as tuned into geometry, or (scary thought) that they expect you to solve the problem through analytical geometry, using a lot of arithmetic.
You could prove ADEF is a parallelogram through pure geometry first, and then you can use the coordinates to prove that ADEF is not a rhombus without much calculation. In fact, you could prove, for a general case, that a parallelogram so constructed is a rhombus if and only if the vertex conserved is the vertex of an isosceles triangle (in other words, if AC=AB).
It all seems obvious and simple in hindsight, and would be easy to explain, except that the "show your work" requirement has to be met following the conventions and formats favored by your teacher, and/or class, and/or textbook.
On the other hand, the arithmetic involved in the analytical geometry approach may be more tolerable than a pure geometry proof in a format that will satisfy the teachers or test graders.
I'll happily leave the choice to you, but I will give you two drawings to refer to.
For pure geometry--> and  <--for analytical geometry
THE GEOMETRY WAY
On connecting the midpoints of the sides of any triangle, you are constructing lines parallel to the sides.
 so since the ratios are the same, lines AC and DE are parallel.
You can prove that the other pairs of lines are parallel in a similar manner.
That way you find out that DE is parallel to AC (or AD), and that FE is parallel to AB (or AF). So ADEF is a parallelogram.
You could also say that you are splitting the original triangle into 4 similar triangles.
AD and AF would be the same length and ADEF would be a rhombus, if and only if AB and AC were the same length, which could happen only if ABC were isosceles.
However, AD and AF are halves of the original triangle sides AC and AD, which are not the same length. So the lengths of AD and AF are not the same, and ADEF is not a rhombus. You prove that AB is longer than AC from the coordinates. The length of horizontal segment AC is the difference of the x coordinates.
AB=
The length of AB could be calculated invoking Pythagoras or the distance formula from the differences of x and y-coordinates for A and B, as
AB=
AB=
I can easily see that
AB= =AC,
and so AB>AC, making AD>AF, and ADEF is not a rhombus.
ONE WAY TO USE ANALYTICAL GEOMETRY
Another way to prove that ADEF is a parallelogram would be to do a lot of analytical geometry calculations from the coordinates of the vertices, calculating midpoints fist. From there you would calculate segment lengths, and/or slopes.
I believe that the easier of the many analytical geometry options would be to show that AF and DE are parallel and have the same length. That makes ADEF a parallelogram. (Other combinations of proving segments are congruent and/or parallel seem to involve more calculations).
The midpoint coordinates are the average of the coordinates for the segment ends, so the coordinates are
for D  , 
for E  ,
for F  , 
So segments DE and AF are part of horizontal lines  and  , and therefore are parallel.
The segments are also congruent, the length of each segment being calculated as the difference of x-coordinates of their endpoints:
DE =  and
AF = 
A pair of congruent and parallel sides (DE and AF) is enough to prove ADEF is a parallelogram.
Then you can calculate the length of AD or EF, and show that it is not also 6, to prove that ADEF is not a rhombus.
EF = 
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