In the triangle $ABC$ $AC=3,BC=4$ and the medians $AA_1$ and $BB_1$ are perpendicular at $M$.
The Pythagorean theorem gives $${AB}^2={AM}^2+{BM}^2\\{A_1B_1}^2={MA_1}^2+{MB_1}^2\\{AB_1}^2={AM}^2+{MB_1}^2\\{BA_1}^2={BM}^2+{MA_1}^2$$ If we multiply the third and fourth equality by $(-1)$ and then add we will get: $$AB^2+A_1B_1^2-AB_1^2-BA_1^2=0$$ From here we can get $AB=\sqrt{5}.$ ($A_1B_1=\dfrac12 AB$) Why can't we just substract the LHSs and RHSs of the equalities to get this equation? I mean can't we say that: $$AB^2-A_1B_1^2-AB_1^2-BA_1^2=AM^2+BM^2-MA_1^2-MB_1^2-AM^2-MB_1^2-BM^2-MA_1^2?$$ Why not? Thank you in advance!
The last equality holds, but that's not what we're searching for. The idea is to get rid of the segments with one endpoint at $H$.