Let be a tetahedron $ ABCD, $ and $ E $ be the projection of $ D $ on the plane formed by $ ABC. $ If $ \mathcal{A}_{\mathcal{R}} $ denotes the area of the region $ \mathcal{R}, $ show that the following affirmations are equivalent: a) $ C=E\vee CE\parallel AB $ b) $ M\in\overline{CD}\implies\mathcal{A}_{ABM}^2=\frac{CM^2}{CD^2}\cdot\mathcal{A}_{ABD}^2 +\left( 1-\frac{CM^2}{CD^2}\right)\cdot\mathcal{A}_{ABC}^2 $