Estonian Math Competitions

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, $r+r_1+r_2=h$.

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Alternative solution.

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 $I_1{K}_{1}D{M}_1$ and $I_2{K}_{2}D{M}_2$ are squares, whereby $D{K}_1=D{M}_1=r_1$ and $D{K}_2=D{M}_2=r_2$. By property of tangent line segments, we have $C{K}_1=C{L}_1$, $AL=AM$ and $A{L}_1=A{M}_1$, whence $$\begin{array}{rl}CD& =C{K}_1+K_{1}D=C{L}_1+r_1=CL+L{L}_1+r_1\\ & =r+AL-A{L}_1+r_1=r+AM-A{M}_1+r_1=r+M{M}_1+r_1.\end{array}$$ By symmetry, we also get $CD=r+M{M}_2+r_2$. After adding up these two equalities and taking into account that $M{M}_1+M{M}_2=D{M}_1+D{M}_2=r_1+r_2$, we obtain $$2CD=2(r+r_1+r_2).$$ Consequently, $CD=r+r_1+r_2$.