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Let ABC be a triangle with \angle BCA=90^{\circ}, and let D be the foot of the altitude from C. Let X be a point in the interior of the segment CD. Let K be the point on the segment AX such that BK=BC. Similarly, let L be the point on the segment BX such that AL=AC. Let M be the point of intersection of AL and BK.

Show that MK=ML.

Proposed by Josef Tkadlec, Czech Republic

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