Show that $AB = 3AD$ 
Given that $AF=EF$ and $BE=CE$. Show that $AB=3AD$.
This question was given during my exams today and it surprised my whole class. No one knew how to start and any tips would be helpful! 

Let $X$ be a point on $BD$ such that $EX\parallel CD$. Then $$\frac{BX}{XD}=\frac{BE}{EC}=1$$ and $$\frac{AD}{DX}=\frac{AF}{FE}=1.$$ It follows that $AD=DX=XB$, so $AB=3AD$.
