Problem 1245

Solution: Let $k$ be the circle with center $D$ and radius 1 . Consider the inversion with respect to the circle $k$ and denote by $A^{*}, B^{*}, C^{*}, A_{1}^{*}, B_{1}^{*}$, and $C_{1}^{*}$ the images of $A, B, C, A_{1}, B_{1}$, and $C_{1}$, respectively.

The point $C_{1}^{*}$ belongs to the line $A^{*} B^{*}$, because the circumcircle of $\triangle A D B$ is mapped to a line. Simiarly, $B_{1}^{*} \in B^{*} C^{*}$ and $A_{1}^{*} \in A^{*} B^{*}$. $D$ belongs to the line $A^{*} A_{1}^{*}$ and $A D=\frac{1}{A^{*} D}$. Using the similar reasoning we get that $D$ is the intersection of $A^{*} A_{1}^{*}, B^{*} B_{1}^{*}$, and $C^{*} C_{1}^{*}$ and the desired equality now becomes equivalent to
$$\frac{A_{1}^{*} D}{A^{*} A_{1}^{*}}+\frac{B_{1}^{*} D}{B^{*} B_{1}^{*}}+\frac{C_{1}^{*} D}{C^{*} C_{1}^{*}}=1 \text {. }$$

Notice that $\frac{A_{1}^{*} D}{A^{*} A_{1}^{*}}=\frac{S_{\triangle B^{*} C^{*} D}}{S_{\triangle B^{*} C^{*} A^{*}}}$. Analogous relations for the remaining two fractions on the left-hand side of (??) further transform our claim to:
$$\frac{S_{\triangle B^{*} C^{*} D}}{S_{\triangle A^{*} B^{*} C^{*}}}+\frac{S_{\triangle C^{*} A^{*} D}}{S_{\triangle A^{*} B^{*} C^{*}}}+\frac{S_{\triangle A^{*} B^{*} D}}{S_{\triangle A^{*} B^{*} C^{*}}}=1,$$
which is obviously true.