ตอนที่ 2 ข้อ 2
กำหนด
ะBAQ=x=
ะQPC,
ะAQR=
ะRQC=y,
ะPAQ=z=
ะACQ,
ะAQB=w ดังนั้นจะได้ว่า
$\displaystyle\frac{\sin y}{RC}=\frac{\sin z}{QR},\quad\frac{\sin y}{AR}
=\frac{\sin z}{QR}
\quad\Rightarrow\quad\frac{AR}{RC}=\frac{\sin z}{\sin x}=:k$
$\displaystyle\frac{\sin z}{PQ}=\frac{\sin (180-w)}{PA}=\frac{\sin w}{PA},
\quad\frac{\sin x}{PQ}=\frac{\sin (2y-w)}{PA}
\quad\Rightarrow\quad\frac{\sin z}{\sin x}=\frac{\sin w}{\sin (2y-w)}$
$\displaystyle\frac{\sin w}{AB}=\frac{\sin x}{QB},\quad\frac{\sin (2y-w)}{BC}
=\frac{\sin z}{QB}
\quad\Rightarrow\quad\frac{AB}{BC}=\frac{\sin w}{\sin (2y-w)}\cdot\frac{\sin z}{\sin x}=k^2$
จบการพิสูจน์
