The triangle midsegment theorem states that the line connecting the midpoints of two sides of a triangle, called the midsegment, is parallel to the third side, and its length is equal to half the length of the third side.
There is a similar theorem for trapezoids: a line connecting the midpoints of the two legs of a trapezoid is parallel to the bases, and its length is equal to half the sum of lengths of the bases.
Problem
ABCD is a trapezoid, AB||CD. EF is a line connecting the midpoints of legs AD and BC, AE=ED and BF=FC. Prove that EF||DC and that EF=½(AB+DC)
Strategy
As we are dealing with the midpoints of segments, we will use what we have already proven for triangle midsegments. Let’s create such triangles, by drawing a line from the vertex A through the midpoint, F, until it intersects an extension of the base DC at point G:
We can easily show that ΔABF and ΔGCF are congruent, using the Angle-Side-Angle postulate. From this, we can show that EF is a midsegment of triangle ΔADG. As such, by the Triangle midsegment theorem, it is parallel to DG and is equal to half of DG. But DG is DC+CG, and as ΔABF and ΔGCF are congruent, CG=AB, so EF is equal to half of DC+AB. In other words, the length of EF is the arithmetic mean (average) of the lengths of the bases.
Proof
(1) AB||DG //Given, ABCD is a trapezoid
(2) ∠BAF ≅ ∠CGF // Alternate Interior Angles Theorem
(3)∠AED ≅ ∠CEF // Vertical angles
(4) BF=FC //Given
(5)ΔABF ≅ ΔGCF // (2), (3), (4), Angle-Side-Angle
(6) AF=FG //(5), corresponding sides of congruent triangles
(7) EF is midsegment //(6), definition of midsegment
(8) EF||DG //(7), Triangle midsegment theorem
(9)EF=½DG //(7), Triangle midsegment theorem
(10) DG = DC+CG
(11) CG=AB //(5), corresponding sides of congruent triangles
(12)EF=½(DC+CG) //(9), (10) , Transitive property of equality
(13) EF=½(DC+AB) //(11), (12) , Transitive property of equality