The Varignon parallelogram of the Varignon parallelogram of quadrilateral ABCD is a 3 x 4 rectangle. Find AC + BD.

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