Question 141
A point $D$ lies inside the triangle $ABC$. If $A_1$, $B_1$, and $C_1$ are the second intersection points of the lines $AD$, $BD$, and $CD$ with the circles circumscribed about $\triangle BDC$, $\triangle CDA$, and $\triangle ADB$ respectively, prove that
$$
\frac{AD}{AA_1} + \frac{BD}{BB_1} + \frac{CD}{CC_1} = 1
$$
$$
\frac{AD}{AA_1} + \frac{BD}{BB_1} + \frac{CD}{CC_1} = 1
$$