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