Problem

Source:

Tags: geometry



In trapezoid $ABCD$ with $AB$ parallel to $CD$ show that : $\frac{|AB|^2-|BC|^2+|AC|^2}{|CD|^2-|AD|^2+|AC|^2}=\frac{|AB|}{|CD|}=\frac{|AB|^2-|AD|^2+|BD|^2}{|CD|^2-|BC|^2+|BD|^2}$