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∆ABC and ∆ADE are similar triangles.
Thus ;
[tex] \frac{AD}{AC} = \frac{DE}{BC} = \frac{AE}{AB} \\ [/tex]
So :
[tex] \frac{5}{3 + 2x} = \frac{4}{8} \\ [/tex]
[tex] \frac{5}{3 + 2x} = \frac{1}{2} \\ [/tex]
Inverse both sides
[tex] \frac{3 + 2x}{5} = \frac{2}{1} \\ [/tex]
[tex] \frac{2x + 3}{5} = 2 \\ [/tex]
Multiply sides by 5
[tex]5 \times \frac{2x + 3}{5} = 5 \times 2 \\ [/tex]
[tex]2x + 3 = 10[/tex]
Thus ;
[tex]AC = 2x + 3 = 10[/tex]
[tex]AC = 10[/tex]
Done...
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