94. From point $A$ at an angle $\alpha$, two rays emerge. On one ray, two points $B$ and $B_{1}$ are taken, and on the other $-C$ and $C_{1}$. Find the length of the common chord of the circumcircles of triangles $A B C$ and $A B_{1} C_{1}$, if $|A B|-|A C|=\left|A B_{1}\right|-\left|A C_{1}\right|=a$.
94. Let (Fig. 23) the second intersection point of the circumcircles of $\triangle A B C$ and $\triangle A B_{1} C_{1}$ be $A_{1}$. From the condition, it follows that $\left|B B_{1}\right|=\left|C C_{1}\right|$, and $\widehat{A B A_{1}}=\widehat{A C A_{1}}$ and $\widehat{A B_{1} A_{1}}=$ $=\widehat{A C_{1} A_{1}}$. Therefore, $\triangle A_{1} B B_{1}=\triangle A_{1} C_{1} C$. Hence, $\left|A_{1} B\right|=$ $=\left|A_{1} C\right|$. Let $\widehat{A B C}=\beta, \widehat{A C B}=\gamma, \widehat{A B A_{1}}=\widehat{A C A_{1}}=\varphi$. Since

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-087.jpg?height=473&width=572&top_left_y=869&top_left_x=336)

Fig. 23.

$\triangle A_{1} B C$ is isosceles, then $\widehat{A_{1} B C}=\widehat{A_{1} C B}$, i.e., $\beta+\varphi=\gamma-\varphi$, $\varphi=\frac{\gamma-\beta}{2}$ and, if the radius of the circumcircle of $\triangle A B C$ is $R$, then $\left|A A_{1}\right|=2 R \sin \frac{\gamma-\beta}{2}$; but $a=: A B^{\prime}-|A C|=$ $=2 R(\sin \gamma-\sin \beta)=4 R \sin \frac{\gamma-\beta}{2} \cos \frac{\beta+\gamma}{2}=2\left|A A_{1}\right| \sin \frac{\alpha}{2}$, therefore, $\left|A A_{1}\right|=\frac{a}{2 \sin \frac{\alpha}{2}}$.