证明$\angle APC=2\angle ABC$
$\therefore$ 三角形内角和为180
$∵
\begin{cases}
∠ABP+∠BAP+∠APB=180
\∠ABC+∠BAC+∠ACB=180
\∠PAC+∠PCA+∠APC=180
\end{cases}
$

$\therefore AP=BP=PC=r$
$\because$△PAB和△PAC为等腰三角形

$\therefore$等腰三角形两底角相等
$\because \angle ABC=\angle BAP=\angle ①,\angle PAC=\angle PCA=\angle ②$

$\because$ 线段BP在线段BC上
$\therefore \angle ABP=\angle ABC=\angle ①$

$∵
\begin{cases}
2∠①+∠APB=180
\∠①+∠②+∠BAC=180
\2∠②+∠APC=180
\end{cases}
$

$\therefore 单位圆所对圆周角90°$
$\because \angle BAC=90°$

$∵
\begin{cases}
2∠①+∠APB=180
\∠①+∠②=90
\2∠②+∠APC=180
\end{cases}
$

$\because 2\angle (90°-\angle ①+\angle APC=180$
$\because 180°-2\angle ①+\angle APC=180$
$\because \angle APC=2\angle ①$

$\because \angle APC=2\angle ABC$