8.5. In an isosceles triangle, the bisector drawn to the base turned out to be twice as short as the bisector drawn to the lateral side. Find the angles of this triangle.

![](https://cdn.mathpix.com/cropped/2024_05_06_4a1cf25cc0b5e0c5ae86g-3.jpg?height=866&width=1153&top_left_y=558&top_left_x=454)
Solution: Let in the isosceles triangle $ABC$ the sides $AB = BC$ and the condition for the bisectors from the problem statement $AE = 2BD$ is satisfied. Extend $ABC$ to the rhombus $ABCF$. Then, since $BD$ is the bisector in the isosceles triangle, it is also a median. Therefore, $BD$ is half of the diagonal of our rhombus, as it intersects the midpoint of the other diagonal $AC$. Thus, points $B, D$, and $F$ lie on the same line.

Notice that $BF = 2BD = AE$, which means the diagonals in the trapezoid $ABEF$ are equal. Therefore, it is an isosceles trapezoid, i.e., $AB = EF$. In particular, triangles $ABE$ and $FEB$ are equal by three sides, from which $\angle BAE = \angle BFE$.

Let $\angle BAE = \angle EAD = \alpha$. Then $\angle FAB = 2 \angle BAC = 4\alpha, \angle FEB = \angle ABE = 180^\circ - 4\alpha, \angle DBC = \frac{1}{2} \angle ABC = 90^\circ - 2\alpha$. Therefore, in triangle $FBE$ from the sum of angles we get

$$
180^\circ = \angle BFE + \angle FEB + \angle FBE = \alpha + (180^\circ - 4\alpha) + (90^\circ - 2\alpha) = 270^\circ - 5\alpha,
$$

from which $\alpha = 18^\circ$, and the angles of the original triangle are $36^\circ, 36^\circ$, and $108^\circ$.

Criteria: Only the answer - 0 points.

The idea to extend to a rhombus - 2 points.