Let $ABC$ be a triangle and denote its area by $k = \mathrm{area}(ABC)$. I want to divide $ABC$ into two sub-triangles $ABE$ and $AEC$ such that $\mathrm{area}(ABE) = t$ and $\mathrm{area}(AEC) = k-t$ for some $t < k$.
Is it possible to find $E$ exactly? By "exactly" I mean that you could get an arbitrarily close approximation by repeatedly bisecting the edge $BC$ to find the split point $E$.
On
$A_{\triangle ABC}=\frac{1}{2}BC\cdot h=k \implies BC=\frac{2k}{h}$ ($h$ is a height from point $A$)
$A_{\triangle ABE}=\frac{1}{2}BE\cdot h=t \implies BE=\frac{2t}{h}$
$A_{\triangle ACE}=\frac{1}{2}(BC-CE)\cdot h=\frac{1}{2}(\frac{2k}{h}-\frac{2t}{h})\cdot h=k-t$
Thus, $BC:BE=k:t$ so $BE=\frac{2t}{h}=\frac{t}{k}BC$ .
I'm not sure what you mean by "compute this exactly". To find the point $E$ on the side $BC$, you can compare the formulae for the areas of the new and original triangles $ABE$ and $ABC$. Assuming $BC$ and $BE$ as the bases in the function $1/2 \times base \times height$, the height remains unchanged and the ratio of the areas will be $t/k$ giving you $BE = BC \times t/k$