In a right triangle \(ACB\) (\(\sphericalangle C = 90^{\circ}\)) \(CH\) is an altitude. On the side \(AC\) we mark a point \(K\) so that \(\sphericalangle CBK = \sphericalangle BAC\). Prove that the line \(CH\) divides the segment \(BK\) in half!

Taisnleņķa trijstūrī \(ACB\) (\(\sphericalangle C = 90^{\circ}\)) novilkts augstums \(CH\). Uz malas \(AC\) atlikts punkts \(K\) tā, ka \(\sphericalangle CBK = \sphericalangle BAC\). Pierādīt, ka taisne \(CH\) dala nogriezni \(BK\) divās vienādās daļās!