# Geometry problems: Area problem…[Homework]

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

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1) Make an affine transformation (this does not affect any of the length ratios) such that $AE$ becomes the bisector of $\hat A$. Since the bisector splits the opposite side in the proportion of the adjacent sides, this will be the case iff $AB=4AC$, i.e. $AD=AC$. Again, because the bisector splits the opposite side in the proportion of the adjacent sides, we conclude $DF=FC$.

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I proved that now It is easy to prove $$S_{ACE} = S_{ADE} = \frac{S_{ABE}}{4}$$, after that, everything is easy too But I haven't done Ex. 2 and 3 yet –  Xeing Nov 15 '12 at 4:08
I finished all of the things now :D, Can't believe it was so easy –  Xeing Nov 15 '12 at 5:32
Ah, great to hear. Indeed, going via the areas is equivalent to though probably less obscure than making the affine transformation argument (after all the theorem about bisectors I used is readily proved via area comparison). –  Hagen von Eitzen Nov 15 '12 at 7:10