The triangles ABC and EDC are similar which means that the rate between corresponding sides is equal, so that:
[tex]\frac{EC}{AC}=\frac{DC}{BC}[/tex]
We know that
EC=7
AC=21
DC=x
BC=20-x
[tex]\begin{gathered} \frac{7}{21}=\frac{x}{20-x} \\ \frac{1}{3}=\frac{x}{(20-x)} \end{gathered}[/tex]
First, you have to multiply both sides by (20-x) to take the x-term from the denominator's place
[tex]\begin{gathered} \frac{1}{3}(20-x)=(20-x)\frac{x}{20-x} \\ \frac{1}{3}(20-x)=x \end{gathered}[/tex]
Next, distribute the multiplication on the parentheses term:
[tex]\begin{gathered} \frac{1}{3}\cdot20-\frac{1}{3}\cdot x=x \\ \frac{20}{3}-\frac{1}{3}x=x \end{gathered}[/tex]
And pass the x-term to the right side of the equation by applying the opposite operation
[tex]\begin{gathered} \frac{20}{3}-\frac{1}{3}x+\frac{1}{3}x=x+\frac{1}{3}x \\ \frac{20}{3}=\frac{4}{3}x \end{gathered}[/tex]
Finally multiply both sides of the expression by the reciprocal fraction of 4/3, i.e. the inverse fraction
[tex]\begin{gathered} \frac{20}{3}\cdot\frac{3}{4}=(\frac{4}{3}\cdot\frac{3}{4})x \\ 5=x \end{gathered}[/tex]
x=5 → so the length of CD is 5 units.
The correct option is D.
