1) $\triangle ABC: D \in AB; E\in BC$ such that $BD=3AD$, $BE = 4EC$. F is the intersection of AE and CD. Prove that FD = FC (I think we should prove $S_{ACE} = S_{ADE} = \frac{S_{ABE}}{4}$)
2)$\triangle ABC: \widehat{A}=90^o, AM \perp BD$ (AM and BD are medians) . $AB = \sqrt{2}$. Find $S_{ABC}$
3)$\triangle ABC: \widehat{A}=30^0; \widehat{B}=50^0$ Prove $ab = c^2 - b^2$ ($\ AB=c; AC=b; BC=a$)
I think we 'll prove $\frac{c+b}{c} = \frac{c}{b} => b(b+c)=c^2$
Please help me, I have more than 10 exercises and have to finish it in one day :pokerface:. Thanks