Let $a = |BC|$, $b = |CA|$, $c = |AB|$ and $h = |CD|$; then $ab = ch = (a + b + c)r$.

Note that the triangles ABC, ACD and CBD are similar by two equal angles (Fig. 28). As the similarity ratio of triangles ACD and ABC is $\frac{b}{c}$ and that of triangles CBD and ABC is $\frac{a}{c}$, we have $r_1 = \frac{b}{c} \cdot r$ and $r_2 = \frac{a}{c} \cdot r$, whence $r + r_1 + r_2 = \frac{a + b + c}{c} \cdot r$. As the equality $ab = (a + b + c)r$ implies $r = \frac{ab}{a + b + c}$, we obtain $r + r_1 + r_2 = \frac{ab}{c}$. On the other hand, the equality $ab = ch$ implies $h = \frac{ab}{c}$. Consequently, $r + r_1 + r_2 = h$.

Let the incenters of the triangles ABC, ACD and BCD be I, I_1 and I_2. Let the points of tangency of the incircle of the triangle ABC with sides BC, CA and AB be K, L and M, respectively; let the points of tangency of the incircle of the triangle ACD with sides DC, CA and AD be K_1, L_1 and M_1, respectively; let the points of tangency of the incircle of the triangle BCD with sides DC, CB and BD be K_2, L_2 and M_2, respectively (Fig. 29). As a tangent line is perpendicular to the radius drawn to the point of tangency, we have $\angle IKC = 90^\circ = \angle ILC$. Thus IKCL is a square by three right angles and the equality $IK = IL$. Hence $CK = CL = r$. Similarly we see that

$I_1K_1DM_1$ and $I_2K_2DM_2$ are squares, whereby $DK_1 = DM_1 = r_1$ and $DK_2 = DM_2 = r_2$. By property of tangent line segments, we have $CK_1 = CL_1$, $AL = AM$ and $AL_1 = AM_1$, whence
$$\begin{aligned} CD &= CK_1 + K_1D = CL_1 + r_1 = CL + LL_1 + r_1 \\ &= r + AL - AL_1 + r_1 = r + AM - AM_1 + r_1 = r + MM_1 + r_1. \end{aligned}$$
By symmetry, we also get $CD = r + MM_2 + r_2$. After adding up these two equalities and taking into account that $MM_1 + MM_2 = DM_1 + DM_2 = r_1 + r_2$, we obtain
$$2CD = 2(r + r_1 + r_2).$$
Consequently, $CD = r + r_1 + r_2$.