Let the area of triangle ODC be x and area of triangle BDC be k. So, the area of triangle OBC = k-x.
In triangles OAB and ODC, $$\angle\ OAB\ =\ \angle\ OCD$$ (Alternate Interior Angles)
Also, $$\angle\ OBA\ =\ \angle\ ODC$$
Hence both the triangles are similar and their linear dimensions are in the ratio 4:5, so their areas will be in the ratio 16:25. Area of triangle OAB = 16x/25. Area of triangle ABC: Area of triangle BDC = 4:5 ( since they have the same height, the area is in the ratio of lengths ). So, [16x/25 + k-x]/k = 4/5. k = 9x/5. Required ratio = (4x/5)/x = 4:5
Alternate Solution:
In triangles OAB and ODC, $$\angle\ OAB\ =\ \angle\ OCD$$ (Alternate Interior Angles)
Also, $$\angle\ OBA\ =\ \angle\ ODC$$
Hence, both the triangles will be similar.
=> $$\frac{AB}{CD}=\ \frac{OB}{OD}=\frac{4}{5}$$ .....(1)
Now construct a perpendicular R from C.
Ratio of area of OBC to area of ODC = $$\ \dfrac{\ area\ of\ OBC}{area\ of\ ODC}\ =\ \ \dfrac{\ \frac{1}{2}\times OB\times\ CR\ }{\frac{1}{2}\times\ OD\times\ CR}=\ \frac{\ OB}{OD}=\ \frac{\ 4}{5}$$
