$\triangle ABC$是等邊三角形,理由如下:
$\because$ 點(diǎn)$A,$$B,$$C,$$P$在$\odot O$上,
$\therefore \angle BAC$與$\angle CPB$都是$\overset{\frown}{BC}$所對(duì)的圓周角,$\angle ABC$與$\angle APC$都是$\overset{\frown}{AC}$所對(duì)的圓周角,
$\therefore \angle BAC=\angle CPB,$$\angle ABC=\angle APC,$
$\because \angle APC=\angle CPB=60^{\circ},$
$\therefore \angle BAC=\angle ABC=60^{\circ},$
$\because$ 在$\triangle ABC$中,$\angle BAC+\angle ABC+\angle ACB=180^{\circ},$
$\therefore \angle ACB=180^{\circ}-\angle BAC-\angle ABC=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ},$
$\therefore \angle BAC=\angle ABC=\angle ACB=60^{\circ},$
$\therefore \triangle ABC$是等邊三角形。