Let $C L$ be the bisector of the external angle $C, D$ be the midpoint of $A B, K$ be a point on the extension of side $A C$ beyond point $C$ (see figure).

First method. Draw a line through vertex $B$ parallel to $C L, F$ be the point of intersection of this line with line $A C$. Then triangle $B C F$ is isosceles. Indeed, $\angle C F B = \angle K C L = \angle B C L = \angle C B F$, so $C F = C B$. Since $A C > B C$, $F$ lies on segment $A C$. Moreover, since $D E \| B F$ and $D$ is the midpoint of $A B$, then $A E = E F$. Thus, $A E = 1 / 2(A C - C F) = 1 / 2(A C - B C) = 1.5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-27.jpg?height=375&width=1083&top_left_y=1822&top_left_x=493)

Second method. According to the theorem on the bisector of the external angle of a triangle, $L A: L B = C A: C B = 8: 5$. By Thales' theorem, $A E: A C = A D: A L = 3: 16$.

Thus, $A E = 3 /{ }_{16} A C = 1.5$.

## Answer

$1.5$.
8,9  [ Properties and characteristics of an isosceles triangle. ]

[ The ratio in which the bisector divides the side ]

In triangle $A B C$, $A C=8, B C=5$. A line parallel to the bisector of the external angle $C$ passes through the midpoint of side $A B$ and point $E$ on side $A C$. Find $A E$.

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