The diagonals of the square ABCD intersect at the point O, SO is the perpendicular to the plane of the square, SO = 4√2. Prove the equality of the catch formed by the lines SA SB SC SD with the plane of the square.
From point S draw straight lines SА, SВ, SC, SD.
Since ABCD is a square, its diagonals AC and BD at point O are divided in half, OA = OA = OC = OD.
SO is perpendicular to the plane of the square, then perpendicular to OA, OB, OS and OD.
Then the triangles SOA, SOB, SОС, SОD are rectangular, in which the SO leg is common, and the second legs are equal, which means that these triangles are equal in two legs.
Then the angle SAO = SBO = SCO = SDO, which was required to prove.
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