I would like to know a better solution for this problem.......'coz I think that my solution is quite weird and quite unreliable , $\frac{[AOD]}{[AOC]}=\frac{OD}{CO}=\frac{5}{9} \implies \boxed{[AOD]=\frac{5}{9} [AOC]}$ Invoking Barycentric Centric Co-ordinates, Let $[ABC]=x$, Let $AB=c , AC=b , BC=a$ $D \equiv (a:b:0) \text{ and } O \equiv (a:b:b)$, Therefore, $$[ADC]=\frac{bx}{a+b} ; [AOC]=\frac{bx}{a+2b} ; [AOD]=\frac{5}{9}(\frac{bx}{a+2b})$$Since, $$[ADC]=[AOD]+[AOC] \implies \frac{bx}{a+b}=\frac{bx}{a+2b}+\frac{5}{9}(\frac{bx}{a+2b}) \implies 9a+18b=14a+14b \implies \boxed{a=\frac{4b}{5}}$$ For the side $AB$ $$AB=c=\sqrt{b^2-a^2}=\frac{3b}{5}$$Sine Rule immediately follows that, $$\sin C=\frac{3}{5} \text{ and } \cos C=\frac{4}{5} \implies \sin \frac{C}{2}=\sqrt{\frac{1-\cos C}{2}}=\frac{1}{\sqrt{10}}$$and $\cos \frac{C}{2}=\frac{3}{\sqrt{10}}$ Now, $$[AOC]=\frac{bx}{a+2b}=\frac{bx}{\tfrac{4b}{5}+2b}=\frac{5x}{14} =\frac{1}{2}9b\sin \frac{C}{2} \implies 5x=63b\sin \frac{C}{2}=\frac{63b}{\sqrt{10}} \implies \boxed{x=\frac{63b}{5\sqrt{10}}}$$ $x=\frac{1}{2}ac=\frac{1}{2}\cdot \frac{4b}{5} \cdot \frac{3b}{5}=\boxed{\frac{6b^2}{25}}$ Hence, $$\frac{63b}{5\sqrt{10}}=\frac{6b^2}{25} \implies \boxed{b=\frac{105}{2\sqrt{10}}}$$ Therefore, $x=\frac{63b}{5\sqrt{10}}=\frac{63\cdot 105}{100}=\boxed{\frac{1323}{20}}$