In both parts, we consider an inversion about $D$ with arbitrary radius $r$. First of all, let us examine what happens to points $A, B, C$ under inversion: \[A^*B^* = \frac{r^2}{DA\cdot DB}AB=\frac{r^2}{DA\cdot DB}\frac{AC\cdot BD}{CD}=\frac{r^2}{AD\cdot CD}AC=A^*C^*.\]Similarly, we may derive that these two lengths are equal to $B^*C^*$. Thus, $\bigtriangleup A^*B^*C^*$ is an equilateral triangle. Next, we look at what happens to the point $C_1$. We know that \[AB\cdot C_1D=AC_1\cdot BD=AD\cdot BC_1.\]Thus, \[\frac{r^2}{DA^*\cdot DB^*} A^*B^*\cdot \frac{r^2}{DC_1^*}=\frac{r^2}{DA^*\cdot DC_1^*}A^*C_1^*\cdot\frac{r^2}{DB^*}=\frac{r^2}{DA^*}\cdot \frac{r^2}{DB^*\cdot DC_1^*}B^*C_1^*\]and simplifying, \[A^*B^*=B^*C_1^*=C_1^*A^*.\]This means that $A^*B^*C_1^*$ is equilateral, and since $C\neq C_1$, thus $C_1^*$ is the reflection of $C^*$ across $A^*B^*$. Finally, we look at what happens to the point $D_1$. We know that \[AB\cdot CD_1=AC\cdot BD_1=AD_1\cdot BC.\]Thus, \[\frac{r^2}{DA^*\cdot DB^*}A^*B^*\cdot\frac{r^2}{DC^*\cdot DD_1^*}C^*D_1^*=\frac{r^2}{DA^*\cdot DC^*}A^*C^*\frac{r^2}{DB^*\cdot DD_1^*}B^*D_1^*=\frac{r^2}{DA^*\cdot DD_1^*}A^*D_1^*\frac{r^2}{DB^*\cdot DC^*}B^*C^*\]and simplifying, \[A^*D_1^*=B^*D_1^*=C^*D_1^*.\]This means that $D_1^*$ is the centre of $\bigtriangleup A^*B^*C^*$. It follows readily that $A^*,B^*,C_1^*,D_1^*$ are concyclic, and hence $A, B, C_1, D_1$ are concyclic. Next, to show that the four points are triharmonic, we wish to show that \[\frac{r^2}{DA_1^*\cdot DB_1^*}A_1^*B_1^*\cdot\frac{r^2}{DC_1^*\cdot DD_1^*}C_1^*D_1^*=\frac{r^2}{DA_1^*\cdot DC_1^*}A_1^*C_1^*\cdot\frac{r^2}{DB_1^*\cdot DD_1^*}B_1^*D_1^*=\frac{r^2}{DA_1^*\cdot DD_1^*}A_1^*D_1^*\cdot\frac{r^2}{DC_1^*\cdot DB_1^*}C_1^*B_1^*.\]But since $D_1^*$ is the centre of $\bigtriangleup A^*B^*C^*$ and hence the centre of $\bigtriangleup A_1^*B_1^*C_1^*$ and $\bigtriangleup A_1^*B_1^*C_1^*$ is equilateral, the conclusion follows readily.