Here, we'll prove the Converse of the Trapezoid Midsegment Theorem - a line that is parallel to the bases of a trapezoid and intersects the midpoint of one of the leg is a midsegment: It intersects the other leg's midpoint, and its length is equal to half the sum of lengths of the bases.
ABCD is a trapezoid. EF is a line segment parallel to its bases, AB and CD, and |AE|=|ED|. Show that EF is a midsegment in the trapezoid, that is, |BF|=|FC| and |EF|=½·(|AB|+|CD|)
We've already proven a similar converse theorem for triangles, so let's try to use the triangle midsegment theorem. For that, we need a triangle - let's create one by drawing the diagonal AC, which intersects EF at point G.
In triangle ΔACD, |EG| is a line parallel to the base CD, which starts from the midpoint of side AD – so by the converse triangle midsegment theorem, it is a midsegment. Because it is a midsegment, |AG|=|GC| and |EG| is equal to half the base. |EG| is then equal to half of CD: |EG|=½·|CD|
Now let’s look at triangle ΔACB. Using the same reasoning as above, because |AG|=|GC|, |FG| starts from the midpoint of side AC and is parallel to AB – so by the converse triangle midsegment theorem, it is a midsegment, and |BF|=|FC|, and we have proven the first part of the converse theorem
But because |FG| is a midsegment, it is also equal to half the base, so |FG|=½·|AB|. Now, |EF|=|EG|+|FG|= ½·|CD|+½·|AB|=½·(|AB|+|CD|), and we have proven the second part.
(1) EG || CD //Given
(2) |AE|=|ED| //Given
(3) EG is a midsegment in ΔACD //(1), (2), converse triangle midsegment theorem
(4) |AG|=|GC| //(3) , defintion of a triangle midsegment
(5) EG || AB //Given
(6) GF is a midsegment in ΔACB //(4), (5), converse triangle midsegment theorem
(7) |BF|=|FC| //(6) , defintion of a triangle midsegment
(8) |EG|=½·|CD| //(3), triangle midsegment theorem
(9) |GF|=½·|AB| //(6), triangle midsegment theorem
(10) |EF|=|EG|+|FG|= ½·|CD|+½·|AB|=½·(|AB|+|CD|)