Answer: Since 1 and 2 have the same measure, angle CED is also equal to 1 and 2. Therefore, triangle CED and triangle CAB are similar by the Angle-Angle (AA) criterion.
Using the properties of similar triangles, we can set up the following proportion:
$\frac{CE}{CA}=\frac{CD}{CB}$
Substituting the given values:
$\frac{CE}{x+4}=\frac{x}{14}$
Cross-multiplying:
$14CE = x(x+4)$
$14CE=x^2+4x$
We also know that triangle ADE and triangle ABC are similar by the AA criterion. Therefore, we can set up the following proportion:
$\frac{DE}{AB}=\frac{AE}{AC}$
Substituting the given values:
$\frac{DE}{18}=\frac{AE}{x+4}$
Cross-multiplying:
$AE = 18\frac{DE}{x+4}$
Now, we can substitute the value of $AE$ in terms of $DE$ into the first equation:
$14CE=x^2+4x$
$14\frac{DE}{x+4}=x^2+4x$
$14DE=x^2+4x(x+4)$
$14DE=x^2+4x^2+16x$
$18x^2+16x-14DE=0$
We can now use the quadratic formula to solve for $x$:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\frac{-16\pm\sqrt{(16)^2-4(18)(-14DE)}}{2(18)}$
$x=\frac{-16\pm\sqrt{256+1008DE}}{36}$
Since $DC=x$, we can now use this equation to find the value of $DC$ for a given value of $DE$. For example, if $DE=5$, we have:
$DC=\frac{-16\pm\sqrt{256+1008(5)}}{36}$
$DC\approx 2.3$ or $DC\approx -3.1$
Since distance cannot be negative, we choose the positive solution:
$DC\approx 2.3$ units.
To find $DE$, we can substitute the value of $DC$ back into one of the earlier equations:
$\frac{CE}{x+4}=\frac{x}{14}$
$\frac{CE}{2.3+4}=\frac{2.3}{14}$
$CE\approx 1.34$ units
Now we can use the second similarity proportion to find $DE$:
$\frac{DE}{18}=\frac{AE}{x+4}$
$\frac{DE}{18}=\frac{18-1.34}{2.3+4}$
$DE\approx 3.64$ units
Therefore, $DC\approx 2.3$ units and $DE\approx 3.64$ units.
Explanation: