I think we have something like this :
IJKL is the 3 x 4 Varignon parallelogram of the Varignon parallelogram of quadrilateral ABCD
IJ =4, IN = OM = 2 and IL = 3
And by the Pythagorean theorem, JL = √[IJ^2 + IL^2] = √[4^2 + 3^2] = √[16 + 9] = √25 = 5 → JM = (1/2)JL = 2.5 = IE
And by the Pythagorean theorem, EM = √[IE^2 − IN^2] = √[2.5^2 − 2^2] = 1.5
So EM = EN + NM = 1.5 + 1.5 = 3 and by symmetry, EG = 2EM = 2*3 = 6 = AD
Similarly......IM = HL = 2.5 and OL = 1.5
So ......HO = √[HL^2 − OL^2] = √[2.5^2 − 1.5^2] = 2
So .....HM = HO + OM = 2 + 2 = 4 and by symmetry, HF = 2HM = 2*4 = 8 = AB
So...by the Pythagorean theorem... AC = BD = √[AD^2 + AB^2] = √[6^2 + 8^2] = √100 = 10
So AC+ BD = 10 + 10 = 20
