Question

find the lengths of the segments with variable expressions.

Explanation:

Step1: Assume similar - triangles

Assume $\triangle ADE\sim\triangle ABC$. Then the ratios of corresponding sides are equal. That is, $\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$. Also, since $EF\parallel BC$, we can use the property of similar - triangles formed by parallel lines.
Let's assume that the lines $AD$, $AE$, $DE$ and $AB$, $AC$, $BC$ are related by the similarity of $\triangle ADE$ and $\triangle ABC$. If we assume that the ratio of similarity is based on the parallel lines $EF\parallel BC$, we know that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$.

Let's assume that the figure is a trapezoid - like structure with parallel lines. We can set up a proportion using the property of similar - triangles formed by parallel lines. If we assume that the two triangles (formed by the non - parallel sides and the parallel lines) are similar, we have:
Let's assume that the ratio of the lengths of the parallel segments is related to the ratio of the lengths of the non - parallel segments.
We know that if we consider the similar triangles formed by the parallel lines $EF\parallel BC$, we can set up the proportion $\frac{x - 4}{2x-7}=\frac{x}{x + x}$.

Cross - multiply:
$(x - 4)\times2x=(2x - 7)\times x$

Step2: Expand both sides

Expand the left - hand side: $(x - 4)\times2x=2x^{2}-8x$.
Expand the right - hand side: $(2x - 7)\times x = 2x^{2}-7x$.
So, $2x^{2}-8x=2x^{2}-7x$.
Subtract $2x^{2}$ from both sides:
$2x^{2}-2x^{2}-8x=2x^{2}-2x^{2}-7x$.
We get $-8x=-7x$.
Add $8x$ to both sides: $-8x + 8x=-7x+8x$.
So, $x = 0$. But this is not a valid solution in the context of lengths (since lengths cannot be negative or zero in this geometric sense).

Let's assume another approach. If we consider the fact that for similar triangles formed by parallel lines in a trapezoid - like figure, we know that $\frac{AD}{AB}=\frac{DE}{BC}$.
Let's assume that the figure is composed of similar triangles. We set up the proportion:
$\frac{x - 4}{2x-7}=\frac{x}{2x}$.
Simplify $\frac{x}{2x}=\frac{1}{2}$.
So, $2(x - 4)=2x-7$.
Expand the left - hand side: $2x-8=2x - 7$.
Subtract $2x$ from both sides: $2x-2x-8=2x-2x - 7$.
We get $-8=-7$, which is a contradiction.

Let's assume that we use the property of similar triangles and set up the proportion based on the fact that if we consider the two similar triangles formed by the parallel lines:
$\frac{AD}{BC}=\frac{AE}{AC}$.
Let's assume that the triangles formed by the non - parallel sides and the parallel lines are similar.
We know that if we consider the parallel lines $EF\parallel BC$, we can set up the proportion $\frac{x - 4}{2x-7}=\frac{x}{2x}$.
Cross - multiplying gives $2x(x - 4)=(2x - 7)x$.
Expanding: $2x^{2}-8x=2x^{2}-7x$.
Subtracting $2x^{2}$ from both sides gives $-8x=-7x$, which is wrong.

Let's assume that we use the mid - segment theorem (if $EF$ is the mid - segment of the trapezoid formed by the parallel lines). If $EF$ is the mid - segment, then $2x= \frac{(x - 4)+(2x - 7)}{1}$.
$2x=x - 4+2x-7$.
$2x=3x - 11$.
Subtract $2x$ from both sides: $0=3x-2x - 11$.
$x = 11$.

Step3: Find the lengths of the segments

For segment $AD=x - 4$, substitute $x = 11$, then $AD=11 - 4=7$.
For segment $BC=2x-7$, substitute $x = 11$, then $BC=2\times11-7=22 - 7 = 15$.

Answer:

$AD = 7$, $BC = 15$