Problem

Source: St Petersburg Olympiad 2010, Grade 11, P5

Tags: geometry, 3D geometry, pyramid



$SABCD$ is quadrangular pyramid. Lateral faces are acute triangles with orthocenters lying in one plane. $ABCD$ is base of pyramid and $AC$ and $BD$ intersects at $P$, where $SP$ is height of pyramid. Prove that $AC \perp BD$