From A and C, drop a perpendicular to BD cutting EF at G and H and BD at I and J, respectively.
Then, by similar triangles AEG = ABI and CFH = CDJ
But, since E anf F are midpoints of AB and CD, and all parts of similar triangles are in the same ratios, EG = 1/2 of BI and FH = 1/2 of JD. Thus (EG + FH) = 1/2 of (BI + JD).
And AC = GH = IJ ...thus GH = 1/2(AC + IJ)
Therefore EF = (EG + FH) + GH = 1/2(BI + JD) + 1/2(AC + IJ) = 1/2(BI + IJ + JD + AC)
But (BI + IJ + JD) = (BD)
So we have
EF = 1/2(BD + AC) = 1/2(AC + BD)
