Diagonals are congruent
|AC| = |BD|
[tex]\dfrac{c}{3}=4-c\ \ \ |multiply\ both\ sides\ by\ 3\\\\\not3^1\cdot\dfrac{c}{\not3_1}=3(4)-3c\\\\c=12-3c\ \ \ |add\ 3c\ to\ both\ sides\\\\4c=12\ \ \ \ |divide\ both\ sides\ by\ 4\\\\\boxed{c=3}\\\\|AC|=|BD|=\dfrac{3}{3}=1[/tex]
