如图,$P$是等边三角形$ABC$外接圆弧$BC$上任意一点,求证:$PA=PB+PC$.<ImageHere>
证明：如答图,在$PA$上截取$PD=PC$,<ImageHere>$becauseAB=AC=BC$,$thereforeangleAPB=angleAPC=60^{circ}$,$thereforetrianglePCD$为等边三角形,$thereforeanglePCD=angleACB=60^{circ},CP=CD$,$thereforeanglePCD-angleDCB=angleACB-angleDCB$,即$angleACD=angleBCP$,在$triangleACD$和$triangleBCP$中,$left{begin{array}{l}AC=BC,angleACD=angleBCP,CD=CP,end{array}right.$$thereforetriangleACDcongtriangleBCP$,$thereforeAD=PB$,$thereforePA=PB+PC$.