Maths Solution
\[
AD + EB + FC = (AB + BC + CD) + (ED + DC + CB) + FC
\]
\[
= AB + (BC + CB) + (CD + DC) + ED + FC
\]
\[
= AB + 0 + 0 + AB + 2AB = 4AB
\]
\[
AD + EB + FC = (AB + BC + CD) + (ED + DC + CB) + FC
\]
\[
= AB + (BC + CB) + (CD + DC) + ED + FC
\]
\[
= AB + 0 + 0 + AB + 2AB = 4AB
\]
