<s>[INST] Quadrilateral/Geometry: Find the area of all the triangles in a trapezoid 
I know the area of $BDE$ is $4$ but can't find out the area of $ABE$.
To solve $BDE$: 
  Since triangle $ACD$ and $CDB$ share the same base and height and knowing their shared area $CDE$ is 3, we know $BDE$ is 4.
Question: How do you solve for $ABE$?  
 [/INST] The area of the triangle ADC is equivalent to area of triangle ACE plus area of triangle CDE $$\frac{1}{2} \times AC \times DC = 4+3$$ $$AC \times DC = 2(4+3).$$ Thus, $AC \times DC = 14.$ 
Now for area of triangle of BDC, it is the area of BDE plus area of CDE $$\frac{1}{2}\times AC\times DC = \text{area of triangle BDE} + 3$$ $$\frac{1}{2}\times 14 = \text{area of triangle BDE} + 3.$$ Thus, area of triangle BDE is 4. 
Area of triangle CDE is the vertical distance of E to DC $\times DC$ such that it is point Y. So, $$3= \frac{1}{2} EY \times DC \text{ so } EY \times DC = 6$$
Furthermore, $$\frac{EY\times DC}{AC\times DC}=\frac{3}{7}.$$ Now, area of ABE is $\frac{1}{2} \times \text{ vertical line from E to arbitrary N}$ such that $$=\frac{1}{2}\times EN\times AB = \frac{1}{2}\times(AC-EY)\times AB$$ $$=\frac{1}{2}\times(AC-\frac{3}{7}AC)\times AB$$ $$=\frac{2}{7}AC\times AB.$$ Now, area of triangle ABC is area of ACE plus area of ABE so $$\frac{1}{2}\times AC\times AB = 4 + \text{area of ABE}$$ $$\frac{1}{2}\times\frac{7}{2}\times\frac{2}{7} \times AC \times AB = 4 + \text{area of ABE}$$ $$\frac{7}{4} \times \text{area of triangle ABE} = 4 + \text{area of ABE}$$ $$\frac{7}{4} \times \text{area of ABE} - \text{area of ABE} = 4$$ $$\frac{3}{4}\times \text{area of ABE} =4.$$ So, $$\text{area of triangle ABE} = \frac{16}{3}.$$ And we are done. 
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