Answer:
[tex]\frac{50}{3}[/tex]
Step-by-step explanation:
∵BC // DE
∴m∠DEC=m∠BCA
∵m∠B=m∠D=90°
∴m∠DCE=m∠BAC
∴ΔCDE similar to ΔABC
∴[tex]\frac{AB}{CD}=\frac{BC}{DE} \\\frac{6}{4}=\frac{8}{DE}[/tex]
∴DE=16/3
[tex]AC=\sqrt{6^{2}+8^{2} } =10[/tex]
[tex]CE=\sqrt{4^{2}+(\frac{16}{3}) ^{2} } =\frac{20}{3}[/tex]
[tex]AE=10+\frac{20}{3} =\frac{50}{3}[/tex]
