?$證明:(1)連接PQ，如圖，$?
?$∵△ABC為等邊三角形，$?
?$∴∠BAC=60°，AB=AC，$?
?$∵線段AP繞點A順時針旋轉(zhuǎn)60°得到線段AQ，$?
?$∴AP=AQ，∠PAQ=60°，$?
?$∴△APQ為等邊三角形，$?
?$∴PQ=AP，$?
?$∵∠CAP+∠BAP=60°，∠BAP+∠BAQ=60°，$?
?$∴∠CAP=∠BAQ，$?
?$在△APC和△ABQ中，$?
?$\begin{cases}{AC=AB}\\{∠CAP=∠BAQ}\\{AP=AQ}\end{cases}$?
?$∴△APC≌△ABQ(\mathrm {SAS})，$?
?$∴PC=QB.$?
?$(2)連接PQ$?
?$∵△ABC為等邊三角形，$?
?$∴∠BAC=60°,AB=AC.\ $?
?$∵線段AP繞點A順時針旋轉(zhuǎn)60°得到線段AQ，$?
?$∴AP=AQ=6,∠PAQ=60°,$?
?$∴△APQ為等邊三角形，$?
?$∴PQ=AP=6.$?
?$∵∠CAP+∠BAP=60°,∠BAP+∠BAQ=60°,$?
?$∴∠CAP=∠BAQ.$?
?$在△APC和△AQB中,$?
?$\begin{cases}{AC=AB}\\{∠CAP=∠BAQ}\\{AP=AQ}\end{cases}$?
?$∴△APC≌△AQB(\mathrm {SAS}),$?
?$∴PC=QB=10.\\ $?
?$∵在△BPQ中,PB2=82,PQ=62,BQ2=102,$?
?$∴PB2+PQ2=BQ ,$?
?$∴△PBQ為直角三角形，∠BPQ=90°,$?
?$∴△APB的高=\frac{1}{2}AP=3$
$∴S△APB=\frac{1}{2}×PB×h=\frac{1}{2}×8×3=12$
