From Angle Bisector to 120 degrees Angle

I'll denote $\angle A=\alpha.$

First off, $\displaystyle S=\frac{1}{2}bc\sin\alpha.\,$ On the other hand, $\displaystyle S=\frac{1}{2}\left(b\cdot l_a\sin\frac{\alpha}{2}+c\cdot l_a\sin\frac{\alpha}{2}\right).\,$ It follows that

$\displaystyle\frac{1}{b}+\frac{1}{c}=\frac{\sin\alpha}{\displaystyle l_a\sin\frac{\alpha}{2}}=\frac{\displaystyle 2\cos\frac{\alpha}{2}}{l_a}.$

Hence, to have $\displaystyle\frac{1}{b}+\frac{1}{c}=\frac{1}{l_a},$ we need $\displaystyle\cos\frac{\alpha}{2}=\frac{1}{2},\,$ i.e. $\displaystyle\frac{\alpha}{2}=60^{\circ}\,$ and $\alpha=120^{\circ}.$

The two other conditions, are somewhat of a red herring, though they do make an elegant problem. In any triangle, $h_a\le l_a\le m_a.\,$ For the second condition, we has, as above, $\displaystyle\cos\frac{\alpha}{2}=\frac{1}{2}\cdot\frac{l_a}{h_a}\ge\frac{1}{2},\,$ making $\alpha\le 120^{\circ}.\;$ Clearly, the less stringent requirement $\displaystyle\frac{1}{b}+\frac{1}{c}\ge\frac{1}{l_a}\,$ would lead to the same conclusion.

The third condition is treated similarly: suffice it to request $\displaystyle\frac{1}{b}+\frac{1}{c}\le\frac{1}{l_a}\,$ to derive $\alpha \ge 120^{\circ}.\;$ The problem would of course loose in elegance.

This is a problem from V. V. Prasolov's Problems in Planimetry v. II, 1986 (in Russian).