Given that the triangle ACD and ABE are similar.
[tex]\Delta ACD\text{ \textasciitilde }\Delta ABE[/tex]
Then their corresponding sides must be proportional;
[tex]\frac{BE}{CD}=\frac{AE}{AD}[/tex]
Given;
[tex]\begin{gathered} BE=20 \\ CD=3x+8 \\ AE=x-1 \\ AD=x-1+27=x+26 \end{gathered}[/tex]
substituting the given values;
[tex]\begin{gathered} \frac{BE}{CD}=\frac{AE}{AD} \\ \frac{20}{3x+8}=\frac{x-1}{x+26} \end{gathered}[/tex]
cross multiply and solve;
[tex]\begin{gathered} 20(x+26)=(3x+8)(x-1) \\ 20x+520=3x^2-3x+8x-8 \\ 20x+520=3x^2+5x-8 \\ 3x^2+5x-8-(20x+520)=0 \\ 3x^2-15x-528=0 \end{gathered}[/tex]
solving the quadratic equation, we have;
[tex]\begin{gathered} 3(x-16)(x+11)=0 \\ so; \\ x=16 \\ or \\ x=-11 \\ \text{ since x cannot be negative then;} \\ x=16 \end{gathered}[/tex]
Therefore, the value of x is;
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