Solution:

A necessary and sufficient condition for $ABCD$ to have an inscribed circle is $AB + CD = BC + AD$. So we have $AB + CD = 2AD$, which we use repeatedly. Extend $DC$ to $X$ so that $BX$ is parallel to $EC$. Then $DX = AB + CD = 2AD$ and the triangles $DEC$, $AEB$, $DBX$ are similar. Let $h$ be the perpendicular distance from $AB$ to $CD$. The similar triangles give us the heights of $DEC$ and $AEB$ in terms of $h$.

$1 / r_1 = \text{perimeter } ABE / (2 \text{ area } ABE) = (AB + 2EB) / (AB \cdot \text{height}) = (AB + 2 \cdot BD \cdot AB / (AB + CD)) / (AB \cdot h \cdot AB / (AB + CD)) = 2(AD + BD)/(AB \cdot h)$. Similarly, $1 / r_3 = 2(AD + BD)/(CD \cdot h)$.

The area of $AED = \text{area } ABD - \text{area } ABE = 1/2 \cdot AB \cdot h \cdot CD / (2AD)$, so $1 / r_2 = 1 / r_4 = \text{perimeter } ADE / (2 \text{ area } ADE) = (AD + BD) / (h \cdot AB \cdot CD / 2AD)$, and $1 / r_2 + 1 / r_4 = 2(AD + BD) / h \cdot 2AD / (AB \cdot CD) = 2(AB + BD) / h \cdot (AB + CD) / (AB \cdot CD) = 1 / r_1 + 1 / r_3$