$x \cos \alpha+y \sin \alpha=\mathrm{k}=x \cos \beta+y \sin \beta$ হলে

$\frac{x}{\cos \frac{1}{2}(\alpha+\beta)}=\frac{y}{\sin \frac{1}{2}(\alpha+\beta)}=?$

দেওয়া আছে,

$\begin{array}{l} x \cos \alpha+y \sin \alpha-\mathrm{k}=0 \\ x \cos \beta+y \sin \beta-\mathrm{k}=0 \end{array}$

বজ্রগুণন প্রক্রিয়ায সাহায্যে (1) ও (2) হতে আমর। পাই , $\frac{x}{-\sin \alpha+\sin \beta}=\frac{y}{-\cos \beta+\cos \alpha}$

$\begin{array}{l} =\frac{k}{\cos \alpha \sin \beta-\sin \alpha \cos \beta} \\ \quad \Rightarrow \frac{x}{2 \cos \frac{1}{2}(\alpha+\beta) \sin \frac{1}{2}(\beta-\alpha)} \\ \quad=\frac{y}{2 \sin \frac{1}{2}(\alpha+\beta) \sin \frac{1}{2}(\beta-\alpha)}=\frac{k}{\sin (\beta-\alpha)} \\ \quad \Rightarrow \frac{x}{2 \cos \frac{1}{2}(\alpha+\beta) \sin \frac{1}{2}(\beta-\alpha)} \\ \quad=\frac{y}{2 \sin \frac{1}{2}(\alpha+\beta) \sin \frac{1}{2}(\beta-\alpha)} \\ \quad=\frac{k \cdot}{2 \sin \frac{1}{2}(\beta-\alpha) \cos \frac{1}{2}(\beta-\alpha)} \\ \therefore \frac{x}{\cos \frac{1}{2}(\alpha+\beta)}=\frac{y}{\sin \frac{1}{2}(\alpha+\beta)}=\frac{k}{\cos \frac{1}{2}(\alpha-\beta)} \end{array}$