Question

in ABCD Trapezium AB is parallelCD,AC an BD intersect in O ,AB=9,CD=6,AC=10,DO=?

Answer 1

Though I can't find the actual length of DO, I'm writing only a way to find DO.

Draw CM, parallel to BO. M will be a point at AB. 
The quadrilateral BMCD thus formed will be a parallelogram. 

∴BM = CD = 6 
AM = AB - BM = 9 - 6 = 3. 

AB || CD. 
∴∠OAB = ∠OCD and ∠OBA = ∠ODC (corresponding angles). 

CM || BO. ∠ACM = ∠AOB and ∠CMA = ∠OBA (corresponding angles). 
∴∠CMA = ∠ODC. 

∴ΔACM ~ ΔAOB ~ ΔODC. 

Sides opposite to equal angles of two similar triangles are in the same ratio. 

∴ On ΔACM ~ ΔODC, 
$\frac{AM}{CD} = \frac{3}{6} = \frac{1}{2} = \frac{AC}{OC} = \frac{CM}{OD}$
AM x 2 = CD = 3 x 2 = 6. 
AC x 2 = OC = 10 x 2 = 20. 
And OD = CM x 2. 

The length of OD must be between 15 to 25. 
Or the average length of OD will be 20.