Ex. 58. In triangle $ABC$, the sides are known: $AB=5, BC=8, AC=7$. On the bisector of angle $A$ (inside the triangle), a point $O$ is chosen such that the areas of triangles $AOB, AOC, BOC$, taken in the given order, form an arithmetic progression. Find this progression.
Ex.58. The semiperimeter $p$ of triangle $ABC$ is 10. Using Heron's theorem, we find that the area of triangle $ABC$ is $S=10 \sqrt{3}$. From the fact that $S_{AOC}$ is the middle term of the progression, it follows that $S_{AOC}=\frac{10 \sqrt{3}}{3}$. Considering that $S_{AOB}: S_{AOC}=5: 7$, we get that $S_{AOB}=\frac{5}{7} \frac{10 \sqrt{3}}{3}=\frac{50 \sqrt{3}}{21}$ and, finally, $S_{BOC}=\left(2-\frac{5}{7}\right) \frac{10 \sqrt{3}}{3}=\frac{90 \sqrt{3}}{21}$. Answer: $\frac{50 \sqrt{3}}{21}, \frac{70 \sqrt{3}}{21}, \frac{90 \sqrt{3}}{21}$.