Prove the ff. theorem. The median of a trapezoid is one-half the sum of the bases. Restatement. Given a trapez

Question

Given a trapezoid ABCD with median segment EF. EF=1/2(AB+CD).

Answer 1

Use basic proportionality theorem. Let AB || CD || EF.
Drop perpendiculars from A to CD and B to CD.
Let they meet EF at G and H & CD at I and J respectively.
Now since AE/AC = 1/2, EG/CI = 1/2
Similarly, FH/DJ = 1/2.
Also AIJB is a rectangle.
Therefore AB = IJ = GH
EF = EG + GH + HF = 1/2 * CI + AB + 1/2 * DJ
Now using the above mentioned identities,, CI + DJ can be written as (CD - AB) / 2
Thus, EF = AB + (CD - AB) / 2 = (AB+CD)/2

Answer 2

Draw vertical line from C and D toward the base and call those intersections with base G and H and intesections with median I and J respectively. Then EI = 1/2 AG and JF = 1/2 HB. Then the median

EF = EI + CD + JF
= 1/2*(AG + HB) + 1/2*2*CD, and since CD = GH
= 1/2*(AG + HB + GH + CD)
= 1/2*(AB + CD).