AoPS - Problem

$H$ being mentioned is purely a red herring. The key observation is that the incenter of $\triangle ABC$ lies on both $\ell_{1}$ and $\ell_{2}$. WLOG let $\ell_{1}$ intersect sides $AB$ and $AC$ at points $C_{1}$ and $B_{1}$, respectively, and the angle bisector of $\angle A$ at $I$. Then denote $d_{1} = \text{dist}(I, \overline{AB}) = \text{dist}(I, \overline{AC})$ as $I$ lies on the $A$-angle bisector and $d_{2} = \text{dist}(I, \overline{BC})$. The equal area condition implies that: \[d_{1}\cdot AB_{1}+d_{1}\cdot AC_{1} = d_{1}\cdot C_{1}B + d_{1}\cdot B_{1}C + d_{2}\cdot BC.\]However, the equal perimeter condition implies that \[AB_{1}+ AC_{1} = C_{1}B + B_{1}C + BC.\]Therefore $d_{2} = d_{1}$, and hence $I$ is the incenter of $\triangle ABC$. Analogously, $I$ lies on $\ell_{2}$ as well. But now $H = \ell_{1}\cap\ell_{2} = I$, so $\triangle ABC$ must be equilateral.