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S^2_{ABM}=CM^2/CD^2 S^2_{ABD}+(1- CM^2/CD^2)S_{ABC}, tetrahedron

Source: IV All-Ukrainian Tournament of Young Mathematicians, Qualifying p11

May 19, 2021
geometry3D geometrygeometric inequalitytetrahedronUkrainian TYM

Problem Statement

In the tetrahedron ABCDABCD, the point EE is the projection of the point DD on the plane (ABC)(ABC). Prove that the following statements are equivalent: a) C=EC = E or CE∥ABCE \parallel AB b) For each point M belonging to the segment CDCD, the following equation is satisfied S△ABM2=CM2CD2⋅S△ABD2+(1−CM2CD2)S△ABC2S^2_{\vartriangle ABM}= \frac{CM^2}{CD^2}\cdot S^2_{\vartriangle ABD}+\left(1- \frac{CM^2}{CD^2} \right)S^2_{\vartriangle ABC} where S△XYZS_{\vartriangle XYZ} means the area of ​​triangle XYZXYZ.
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