In triangle $ABC$, side $AB=15$ and $AC=10$, $AD$ is the angle bisector of angle $A$. A line is drawn from point $D$, parallel to $AB$, until it intersects $AC$ at point $E$. Find $AE$, $EC$, and $DE$.

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The bisector of a triangle divides its side into segments proportional to the other two sides.

## Solution

Since the bisector of a triangle divides its side into segments proportional to the other two sides, then $C D: D B=A C: A B=2: 3$.

Therefore, $D E=2 / 5 A B=6$ (since triangles $C D E$ and $C B A$ are similar). By Thales' theorem, $A E=3 / 5 A C=6$, $E C=A C-A E=10-6=4$.

## Answer

$6,4,6$.