[ Two pairs of similar triangles]

In triangle $ABC$, point $D$ bisects side $AB$, and point $E$ lies on side $BC$, such that segment $BE$ is one-third the length of side $BC$. Segments $AE$ and $CD$ intersect at point $O$. Find $AB$, given that $AE=5$, $OC=4$, and $\angle AOC=120^{\circ}$.
Continue the segment $A E$ until it intersects with the line drawn through vertex $C$ parallel to $A B$, and consider two pairs of similar triangles formed.

## Solution

Draw a line through point $C$ parallel to $A B$, and extend the segment $A E$ until it intersects the drawn line at point $T$. From the similarity of triangles $C E T$ and $B E A$, it follows that $C T = 2 A B = 4 A D$.

From the similarity of triangles $C O T$ and $D O A$, we find that $C O: O D = T O: O A = C T: A D = 4: 1$. Therefore, $O D = \frac{1}{4} C O = 1$,

$A O = \frac{1}{5} A T = \frac{1}{5}(A E + E T) = \frac{3}{5} A E = 3$.

By the cosine rule, $A D^{2} = A O^{2} + D O^{2} - 2 A O \cdot D O \cos \angle A O D = 9 + 1 - 2 \cdot 3 \cdot 1 \cdot \cos 60^{\circ} = 7, A B = 2 A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-12.jpg?height=429&width=778&top_left_y=1100&top_left_x=640)

## Answer

$2 \sqrt{7}$.