Let triangle ABC be a triangle and let D, E, an F be the midpoints of the sides BC, AC, and AB, respectively
a) Prove that triangle EDC is not similar to triangle ABC.
b) Prove that the congruences segment AF congruent to segment ED, segment AE congruent segment FD, and segment BD congruent to segment EF cannot all hold.
Angle CED is congruent to angle CAB (corresponding angles)
Angle CDE is congruent to angle CBA (corresponding angles)
Therefore, triangle EDC is similar to triangle ABC because of AA similarity property.
You can answer part (b) by what I stated about the midsegment of a triangle. The midsegment of a triangle is the segment that joins the midpoints of two sides. This segment is parallel to the third side and is one-half the length of the third side. This is enough information to show that
ED = 1/2 AB (triangle midsegment theorem)
AF = 1/2 AB (Given)
Therefore, AF = ED
Using similar logic, you can prove AE = FD and BD = EF.