We are given that the lines $(\ell_1)$ and $(\ell_2)$ pass through the orthocenter $H$ of triangle $ABC$ and each line divides the triangle into two figures of equal area and equal perimeters. We need to determine the angles of the triangle.

The orthocenter $H$ of a triangle is the intersection of its altitudes. For the line $(\ell_1)$ to divide the triangle $ABC$ into two parts of equal area, it must pass through $H$ and reach the midpoints of the sides of the triangle. Similarly, line $(\ell_2)$ must satisfy the same condition. If both lines divide the triangle into regions of equal perimeter as well as equal area, this implies symmetry.

For the configuration where such conditions hold, consider an equilateral triangle:

1. In an equilateral triangle with sides $a$, all altitudes are equal, and the medians and altitudes coincide. The orthocenter $H$ is the same as the centroid and the circumcenter.

2. If line $(\ell_1)$ passes through $H$, it can align with any median (which is also an altitude). Given the symmetry, the division will always result in parts with equal area and perimeter.

3. Similarly, line $(\ell_2)$ can align with another median. In an equilateral triangle, any line through the orthocenter divides the triangle into regions of equal area and perimeter due to its symmetry.

When equilateral conditions are not satisfied, such divisions generally do not hold, as the perimeters of resulting sections would differ once they form different shaped sections other than those symmetric to each other, distinct in non-equilateral triangles.

Therefore, the only triangle for which two such lines exist is an equilateral triangle. Thus, every angle in the triangle must be:
$$\boxed{60^\circ}$$

Hence, the angles of the triangle are $60^\circ, 60^\circ, 60^\circ$.