Bay Area Mathematical Olympiad

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 $\triangle ADB$ is mapped to a line. Similarly, $B_1^* \in B^* C^*$ and $A_1^* \in A^* B^*$. $D$ belongs to the line $A^* A_1^*$ and $AD = \frac{1}{A^* D}$. Using 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
$$\frac{A_1^* D}{A^* A_1^*} + \frac{B_1^* D}{B^* B_1^*} + \frac{C_1^* D}{C^* C_1^*} = 1$$
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 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.