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A Nice Little Area Ratio Lemma

Observe the high quality diagram above. We see that $\triangle{AEB}$ and $\triangle{BEC}$ share the same height (that height being the same height of $\triangle{ABC}$). Let the height be $h$. If $[AEB]$ denotes the area of $\triangle{AEB}$ then

$\dfrac{[AEB]}{[BEC]} = \dfrac{AE*\frac{h}{2}}{AC*\frac{h}{2}} = \dfrac{AE}{AC}$. Thus $[AEB]:[BEC] = AE:AC$

Just like $\triangle{AEB}$ and $\triangle{EBC}$ share a common height the same is true for $\triangle{AED}$ and $\triangle{EDC}$. And what's even more convenient is the fact that if we call the common height between $\triangle{AED}$ and $\triangle{EDC} h_1$ we get that

$\dfrac{[AED]}{[EDC]} = \dfrac{AE*\frac{h_1}{2}}{EC*\frac{h_1}{2}} = \dfrac{AE}{EC} = \dfrac{[ABE]}{[BEC]}$!

Here's a quick algebra fact: If $\frac{a}{b} = \frac{c}{d}$ then $\frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d} = \frac{a-c}{b-d}$.

From $\dfrac{[AEB]}{[BEC]} = \dfrac{[AED]}{[EDC]} = \dfrac{[AEB]-[AED]}{[BEC]-[EDC]} = \dfrac{[BDA]}{[BDC]} = \dfrac{AE}{EC}$ we obtain that

$[ADB]:[BDC] = AE:EC$ which is slightly less trivial than the other ratios stated above.