If ABCD is a trapezium, AC and BDarethe diagonals intersecting each otherat point O.
Then AC : BD is

Given, in trapezium ABCD, 

$$AB\parallel DC$$

in $$\triangle OAB$$ and $$\triangle OCD$$

$$\angle OAB=\angle OCD$$ (internal alternate angles)

$$\angle OBA=\angle ODC$$ (internal alternate angles)

$$\angle AOB=\angle COD$$ (vertically opposite angles)

therefore by AAA similarity, $$\triangle OAB\sim\triangle OCD$$

by using triangle similarity property

$$\frac{OA}{OC}=\frac{OB}{OD}$$       $$- equation 1$$

by subtracting 1 from both sides, we get

$$\frac{OA}{OC}-1=\frac{OB}{OD}-1$$

$$\frac{OA-OC}{OC}=\frac{OB-OD}{OD}$$

$$\frac{OA-OC}{OB-OD}=\frac{OC}{OD}$$    $$- equation 2$$

by adding 1 to both sides in equation 1, we get

$$\frac{OA}{OC}+1=\frac{OB}{OD}+1$$

$$\frac{OA+OC}{OC}=\frac{OB+OD}{OD}$$

$$\frac{AC}{BD}=\frac{OC}{OD}$$      $$- equation 3$$

From equations $$2,3$$

$$\frac{AC}{BD}=\frac{OA-OC}{OB-OD}$$