Calculate AE in Triangle: Parallel Lines with Segments 6 and 10

Let's prove that triangles ADE and ABC are similar using:

Since DE is parallel to BC, angles ADE and ABC are equal (according to the law - between parallel lines, corresponding angles are equal)

Angle DAE and angle BAC are equal since it's the same angle

After we proved that the triangles are similar, let's write the given data from the drawing according to the following similarity ratio:

$\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$

We know that - $AB=AD+DB=6+4=10$

$\frac{6}{10}=\frac{10}{154}=\frac{AE}{AC}$

Let's reduce the fractions:

$\frac{3}{5}=\frac{2}{3}$

This statement is incorrect, meaning the data in the drawing contradicts the fact that the triangles are similar. Therefore, the drawing is impossible.