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midpoint of edge equidistant from vertices in triangular pyramid,Caucasus 11th

Source: I Caucasus 2015 11.4

September 6, 2018
geometry3D geometrypyramid3-Dimensional Geometry

Problem Statement

The midpoint of the edge SASA of the triangular pyramid of SABCSABC has equal distances from all the vertices of the pyramid. Let SHSH be the height of the pyramid. Prove that BA2+BH2=CA2+CH2BA^2 + BH^2 = C A^2 + CH^2.
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