Part A
[tex]\overline{AC}[/tex] and [tex]\overline{BD}[/tex]
Part B
Diagonals of a parallelogram bisect each other, so
[tex]BE=ED \implies 2x+3=3y-4 \implies 3y=2x+7 \implies y=\frac{2}{3}x+\frac{7}{3}\\\\AE=EC \implies y+4=4x-6 \implies y=4x-10[/tex]
Solving by substitution,
[tex]\frac{2}{3}x+\frac{7}{3}=4x-10[/tex]
Multiplying both sides by 3,
[tex]2x+7=12x-30\\\\37=10x\\\\x=3.7[/tex]
So, [tex]y=\frac{2}{3}\left(\frac{37}{10} \right)+\frac{7}{3}=\frac{24}{5}[/tex]
Part C
[tex]BE=ED=2(3.7)+3=10.4\\\\AE=EC=4(3.7)-6=8.8[/tex]
[tex]\therefore BD=2(10.4)=20.8, AC=2(8.8)=17.6[/tex]
