Prove that line which connects the intersection point of trapezoid legs' extensions and the intersection point of its diagonals divides its bases in half. Mathematically, prove that: \(\displaystyle |DE|=|CE|=\dfrac{1}{2}|CD| \, \text{and} \, |AF|=|BF|=\dfrac{1}{2}|AB| \). I don't know how to start here.
But, I thought, \(\displaystyle |GS_2| \) is the altitude on \(\displaystyle \overline{AB} \) of \(\displaystyle \bigtriangleup ABS_2 \). The area of \(\displaystyle \bigtriangleup ABS_2 \) is: \(\displaystyle A(ABS_2)=\dfrac{1}{2}|GS_2||AB| \). But \(\displaystyle |GS_2| \) is also the altitude of \(\displaystyle \bigtriangleup AFS_2 \) on \(\displaystyle \overline{AF} \) and of \(\displaystyle \bigtriangleup BFS_2 \) on \(\displaystyle \overline{BF} \). Both triangles have the same altitude \(\displaystyle |GS_2| \) and since \(\displaystyle |AF|=|BF| \), they also have the same areas: \(\displaystyle A(AFS_2)=\dfrac{1}{2}|GS_2||AF| \) and \(\displaystyle A(BFS_2)=\dfrac{1}{2}|GS_2||BF| \). If I could somehow prove now that \(\displaystyle A(AFS_2)=A(BFS_2) \), then, since they have the same altitude, I would prove that \(\displaystyle |AF|=|BF| \). But, I don't know how to do that and nothing else comes to my mind.