Let $k = k(O, r)$ and let $k_1(I, r_1)$ touch $BD$, $CD$ and $k$ at $P$, $R$ and $Q$, respectively. Let $k_2(J, r_2)$ touch $AD$, $CD$ at $K$, $M$ and $L$, respectively. First, we shall prove that $AB \parallel IJ$, i.e. $ABIJ$ is an isosceles trapezoid. Assume the contrary and set $T = IJ \cap AB$. Then

$$\frac{IT}{TJ} \cdot \frac{JL}{LO} \cdot \frac{OQ}{QI} = \frac{r_1}{r_2} \cdot \frac{r_2}{r} \cdot \frac{r}{r_1} = 1$$

and, by the Menelaus theorem, the points $T$, $Q$ and $L$ are collinear.

Then $TQ \cdot TL = TB \cdot TA$. On the other hand, $ABIJ$ is cocyclic which implies $TB \cdot TA = TI \cdot TJ$. It follows that $TQ \cdot TL = TI \cdot TJ$, i.e. $QLJI$ is cocyclic. But $\nparallel JLQ = \nparallel IQL$, i.e. $QLJI$ is an isosceles trapezoid and then $IJ \parallel QL$, a contradiction. Hence $AB \parallel IJ$, i.e. $r_1 = r_2$ and $AK = BP$ (1).

Further, the generalized Ptolemy theorem (applied to $A$, $B$, $Q$ and $C$) gives $AB \cdot CR + AC \cdot BP = AP \cdot BC$. Since $CR = CD - DR = CD - DP = CD - BD + BP$ and $AP = AB - BP$, we get

$$(2) \quad BP = \frac{AB(BC + BD - CD)}{AB + BC + AC}$$

Analogously,

$$(3) \quad AK = \frac{AB(AC + AD - CD)}{AB + BC + AC}$$

Finally, (1), (2) and (3) imply $BC + BD = AC + AD$ which holds if and only if $D$ is the tangent point of $AB$ and the excircle to this side.