I don't get calc rulz's solution. I'd think the area in that case would be $ \frac {2}{9}$ of the original area. Also, with ociretsih's solution, $ \frac{2b}{b+c}$ is not symmetric with respect to any two variables, so how can that be the area? [geogebra]5ce70bf2b979a6c6d19d5c929e964903f417c34e[/geogebra]

Hmm I will clarify ociretsih's solution. Assume without loss of generality that $AB\ge AC\ge BC$. We wish to show that if we reflect about the angle bisector of $\angle{BAC}$ (this is motivated by the isosceles case), \[\frac{[\triangle{ABC}\cap\triangle{AB'C'}]}{[ABC]} = 1-\frac{[BC'D]}{[ABC]} > \frac23.\]We can see that \begin{align*} \frac{[BC'D]}{[ABC]} = \frac{BC'}{BA}\cdot\frac{BD}{BD+DC} = \frac{BC'}{BA}\cdot\frac{1}{1+\frac{CD}{BD}} = \frac{BC'}{BA}\cdot\frac{1}{1+\frac{CC'}{BB'}} &= \frac{BC'}{BA}\cdot\frac{1}{1+\frac{C'A}{BA}}\\ &= \frac{BC'}{BA+C'A} = \frac{BC'}{BC'+2C'A}, \end{align*}so it is equivalent to show that \[1-\frac{BC'}{BC'+2C'A} > \frac23 \Longleftrightarrow AC=AC'>BC'.\]The Triangle Inequality gives us \[BC+AC > AB = AC'+BC' = AC+BC' \Longleftrightarrow BC>BC'.\]Since $AC\ge BC$, $AC\ge BC>BC'$ so we're done. Edit: Whoops if you want you can just use the Angle Bisector Theorem to find the ratios, but I forgot about it.

calc rulz solution is incorrect since the problem specifically asked for a line l such that the area is strictly greater than 2/3.

In fact, we can tighten this bound a little bit. Claim: If $a \le b \le c$, then $\frac{b}{a} < \frac{1+\sqrt{5}}{2}$ or $\frac{c}{b} < \frac{1+\sqrt{5}}{2}$. Proof: Assume not: then $c-b \ge \frac{\sqrt{5}-1}{2}b \ge a$, violating the triangle inequality. Now, if $\frac{b}{a} < \frac{1+\sqrt{5}}{2}$, then we take the bisector of angle $C$, giving us $\frac{2a}{a+b} > \frac{2}{1+\frac{1+\sqrt{5}}{2}}=3-\sqrt{5}$. If $\frac{c}{b} < \frac{1+\sqrt{5}}{2}$, then we take the bisector of angle $A$, giving us $\frac{2b}{b+c} > \frac{2}{1+\frac{1+\sqrt{5}}{2}}=3-\sqrt{5}$. However, $3-\sqrt{5} \approx 0.763932023 > \frac{2}{3}$.

All that's needed is: Claim: If $ABC$ is a triangle where $\frac{1}{2} < \frac{AB}{AC} < 1$, then the $\angle A$ bisector works. Proof. Let the $\angle A$-bisector meet $BC$ at $D$. The overlapped area is $2[ABD]$ and \[ \frac{[ABD]}{[ABC]} = \frac{BD}{BC} = \frac{AB}{AB+AC} \]by angle bisector theorem. $\blacksquare$ In general, suppose $x < y < z$ are sides of a triangle. Then $\frac{1}{2} < \frac yz < 1$ by triangle inequality as needed.

Finding such a line $l$ for an equilateral triangle $\Delta$ is trivial. To solve it in the general case, let $\varphi$ be an affine transformation taking $\Delta \mapsto \triangle ABC$ (which always exists), so $\varphi: \Delta\cup l\to \triangle ABC\cup \varphi(l)$. Since afine transformations conserve area ratios, and ratios on segments (and hence $\varphi$ commutes with reflection over $l$ or its image), the line $\varphi(l)$ solves the problem for $\triangle ABC$.

Assume WLOG that $AB$ = 1, because the problem is unaffected by scaling. Now, at least one of ${AC, BC}$ lies in $\left(\frac{1}{2}, 2\right)$. It's easy to show this is true by the triangle inequality. WLOG $AC \in \left(\frac{1}{2}, {2}\right)$ and further suppose $AC \geq AB$, so $AC \in [1, 2)$. Then, let the A-bisector hit $BC$ at $D$, and reflect over this bisector. Then $AB' > B'C$ so $[AB'D] > [CB'D]$. It follows that the desired area (the intersection of $ABC$ and $A'B'C'$) equals $[ABD] + [AB'D]$, and $$[ABD] + [AB'D] = 2[AB'D] = \frac{(2[AB'D])(2[AB'D] + [CB'D])}{2[AB'D] + [CB'D]} > \frac{(2[AB'D])(2[AB'D] + [CB'D])}{3[AB'D]} = \frac{2[ABC]}{3}$$so we're done.

If WLOG $BC \le CA \le AB$, then reflecting about the angle bisector of $\angle BAC$ suffices; the condition is equivalent then to $AB \le AC \le 2AB$ which is true. $\square$