1. Define the given elements and recall known formulas:
- Let $S = S(ABCD)$ be the area of the tangential trapezoid $ABCD$.
- Let $S_1 = S(\triangle ABE)$, $S_2 = S(\triangle BCE)$, $S_3 = S(\triangle CDE)$, and $S_4 = S(\triangle DAE)$ be the areas of the triangles formed by the diagonals and the vertices of the trapezoid.
- Let $s_1 = \frac{\text{Perimeter}(\triangle ABE)}{2}$, $s_2 = \frac{\text{Perimeter}(\triangle BCE)}{2}$, $s_3 = \frac{\text{Perimeter}(\triangle CDE)}{2}$, and $s_4 = \frac{\text{Perimeter}(\triangle DAE)}{2}$ be the semiperimeters of these triangles.
- Given that $AD \parallel BC$.

2. Recall the well-known formulas:
- The area of a triangle with an inscribed circle is given by $S = r \cdot s$, where $r$ is the inradius and $s$ is the semiperimeter.
- For the tangential trapezoid, the sum of the lengths of the opposite sides is equal: $AB + CD = AD + BC$.

3. Use the similarity of triangles:
- Since $\triangle DAE \sim \triangle BCE$, we have:
$$\frac{S(\triangle DAE)}{S(\triangle BCE)} = \left( \frac{DA}{BC} \right)^2 = \left( \frac{AE}{CE} \right)^2 = \left( \frac{ED}{EB} \right)^2$$

4. Express the given problem in terms of the inradii:
- We need to prove that:
$$\frac{1}{r_1} + \frac{1}{r_3} = \frac{1}{r_2} + \frac{1}{r_4}$$
- Using the formula $S = r \cdot s$, we can write:
$$\frac{1}{r_1} = \frac{s_1}{S_1}, \quad \frac{1}{r_2} = \frac{s_2}{S_2}, \quad \frac{1}{r_3} = \frac{s_3}{S_3}, \quad \frac{1}{r_4} = \frac{s_4}{S_4}$$

5. Simplify the expressions:
- Given that $S_1 = S_3 = \sqrt{S_2 S_4}$, we can write:
$$\frac{1}{r_1} + \frac{1}{r_3} = \frac{s_1}{S_1} + \frac{s_3}{S_3} = \frac{s_1}{\sqrt{S_2 S_4}} + \frac{s_3}{\sqrt{S_2 S_4}} = \frac{s_1 + s_3}{\sqrt{S_2 S_4}}$$
- Similarly, we have:
$$\frac{1}{r_2} + \frac{1}{r_4} = \frac{s_2}{S_2} + \frac{s_4}{S_4} = \frac{s_2}{\sqrt{S_2 S_4}} + \frac{s_4}{\sqrt{S_2 S_4}} = \frac{s_2 + s_4}{\sqrt{S_2 S_4}}$$

6. Use the property of the tangential trapezoid:
- Since $AB + CD = AD + BC$, the semiperimeters satisfy:
$$s_1 + s_3 = s_2 + s_4$$

7. Conclude the proof:
- Therefore, we have:
$$\frac{1}{r_1} + \frac{1}{r_3} = \frac{s_1 + s_3}{\sqrt{S_2 S_4}} = \frac{s_2 + s_4}{\sqrt{S_2 S_4}} = \frac{1}{r_2} + \frac{1}{r_4}$$

$\blacksquare$