Answer to Question #223144 in Geometry for Mac Evans

Question

Consider a parallelogram ABCD. Show that AC² + BD² = AB² + BC² + CD² + AD²

Answer 1

Let us consider a parallelogram $ABCD$ and show that $AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + AD^2$.

According to Cosine Theorem, $AC^2=AB^2+BC^2-2AB\cdot BC\cos\angle ABC$ and $BD^2=BC^2+CD^2-2BC\cdot CD\cos\angle BCD.$

Taking into acount then in the parallelogram $ABCD$ the angles $\angle ABC$ and $\angle BCD$ are supplementary, and $AB=CD,\ BC=AD,$ we conclude that

$AC^2+BD^2=(AB^2+BC^2-2AB\cdot BC\cos\angle ABC)+(BC^2+CD^2-2BC\cdot CD\cos\angle BCD)\\ =AB^2+BC^2-2AB\cdot BC\cos\angle ABC+AD^2+CD^2-2BC\cdot AB\cos(180^{\circ}-\angle ABC)\\ =AB^2+BC^2+CD^2+AD^2-2AB\cdot BC\cos\angle ABC+2BC\cdot AB\cos(\angle ABC)\\ =AB^2+BC^2+CD^2+AD^2.$