5. In triangle $ABC$, $|AB|=30 \, \text{mm}, |AC|=60 \, \text{mm}$. From point $D$ on side $\overline{AC}$, a line is drawn that intersects side $\overline{AB}$ at point $E$ such that $|\measuredangle ADE|=|\measuredangle CBA|$. Determine $|AD|$ and $|AE|$ if $|AE|$ is $6 \, \text{mm}$ longer than $|AD|$.

Problems worth 10 points:
First method:

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_494875a1129ddb904645g-23.jpg?height=403&width=362&top_left_y=381&top_left_x=356)

Angles $\measuredangle A D E$ and $\measuredangle C B A$ are congruent, and angles $\measuredangle E A D$ and $\measuredangle B A C$ are congruent (common), so by the AA similarity theorem, triangles $A D E$ and $A B C$ are similar.

If the triangles are similar, the lengths of corresponding sides are in the same ratio:

$|A E|:|A D|=|A C|:|A B|=2: 1$ thus,

1 POINT

$|A E|=2|A D|$.

1 POINT

From the condition of the ratio of the lengths of sides $|A D|$ and $|A E|$, it follows that

$|A E|=|A D|+6$

$2|A D|=|A D|+6$

1 POINT

$|A D|=6 \mathrm{~mm},|A E|=12 \mathrm{~mm}$

1 POINT