Consider ΔACD.
Line-segment CD is bisected by AB at O. Therefore, AO is the median of ΔACD. ∴Area(ΔACO)=Area(ΔADO)...(1)
Considering ΔBCD,BO is the median.
∴Area(ΔBCO)=Area(ΔBDO)...(2)
Adding equations(1) and (2), we obtain;
Area(ΔACO)+Area(ΔBCO)=Area(ΔADO)+Area(ΔBDO)
⇒Area(ΔABC)=Area(ΔABD)