$AH$ is the altitude of the acute triangle $ABC$, $K$ and $L$ are the feet of the perpendiculars, from point $H$ on sides $AB$ and $AC$ respectively. Prove that the angles $BKC$ and $BLC$ are equal.
Since $\angle{AKH} =\angle{ALH} =90 \Longrightarrow AKHL$ is a cyclic quadrilateral.
$$\Longrightarrow \angle{AKL} =\angle{AHL} =90-\angle{LHC} =\angle{C} $$$\Longrightarrow KLCB$ is a cyclic quadrilateral $\Longrightarrow \angle{BKC} =\angle{BLC}. $
As desired...
$\blacksquare$