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Firstly, we consider the case when point $L$ belongs to the segment $AH$ (fig. 21). Since $\angle LHB = \angle LDB = 90^\circ$, then $BLHD$ is cyclic quadrilateral. Thus, $\angle LBD = \angle DHC$. On the other hand $\angle LBD = \angle LBA$, due to the fact that $BL$ is a bisector and $\angle DHC = \angle BAC$, because $AB \parallel HD$. It follows that $\angle ABC = 2\angle BAC$ and $\angle ABC + \angle BAC = 180^\circ - 60^\circ = 120^\circ$. And we get $\angle BAC = 40^\circ$ as $\angle ABC = 80^\circ$.

We point out that $\angle LBD = \angle DHL$ and again $\angle LBD = \angle LBA$, due to the fact that $BL$ is a bisector of the triangle and $\angle DHL = \angle BAC$ since $AB \parallel HD$. Similarly we get $\angle BAC = 40^\circ$ and $\angle ABC = 80^\circ$. But, if $\angle ACB = 60^\circ > \angle BAC = 40^\circ$ then the point $H$ can not belong to the segment $AL$. Thus, this case is impossible.