Circles of Apollonius of a Triangle 2

Let the sides of the triangle be $a$ , $b$ , and $c$ , let the angles be $\angle A$ , $\angle B$ , and $\angle C$ , where $\angle C$ is the obtuse angle, and let $D$ be the intersection of the three circles on side $c$ , where $d = CD$ , as follows:

Then $\frac{a}{b} = \frac{BD}{AD}$ , which by the angle bisector means that $\angle BCD = \angle ACD$ . Let $\theta = \angle BCD = \angle ACD$ .

Also since $\frac{a}{b} = \frac{BD}{AD}$ , then $\frac{a}{b} = \frac{BD}{c - BD}$ , which solves to $BD = \frac{ac}{a + b}$ , and $AD = c - CD = c - \frac{ac}{a + b}$ , which solves to $AD = \frac{bc}{a + b}$ .

Also, $\frac{a}{c} = \frac{d}{AD}$ , and since $AD = \frac{bc}{a + b}$ , this solves to $d = \frac{ab}{a + b}$ .

Then by the law of cosines on $\triangle BCD$ , $(\frac{ac}{a + b})^2 = a^2 + (\frac{ab}{a + b})^2 - 2 \cdot a \cdot \frac{ab}{a + b} \cos \theta$ , which simplifies to $c^2 = (a + b)^2 + b^2 - 2(a + b)b \cos \theta$ , and by the law of cosines on $\triangle ACD$ , $(\frac{bc}{a + b})^2 = b^2 + (\frac{ab}{a + b})^2 - 2 \cdot b \cdot \frac{ab}{a + b} \cos \theta$ , which simplifies to $c^2 = (a + b)^2 + a^2 - 2(a + b)a \cos \theta$ .

Therefore, $c^2 = (a + b)^2 + b^2 - 2(a + b)b \cos \theta = (a + b)^2 + a^2 - 2(a + b)a \cos \theta$ , which rearranges to $\cos \theta = \frac{1}{2}$ , so $\theta = 60°$ and $\angle C = 2\theta = \boxed{120°}$ .