P是等边三角形ABC内一点,连接PA,PB,PC,以BP为边做角PBQ等于六十度且BQ=BP,连接CQ,观察AC与CQ大

）相等∵等边△ABC∴AB = BC,∠ABC = 60°∵∠PBQ = 60°∴∠ABP = ∠CBQ∵BP = BQ∴△ABQ≌△CBQ∴AP = CQ2）直角三角形证明：∵∠PBQ = 60°,BP = BQ∴△BPQ是等边三角形∴PQ = BP ∵AP = CQ（第一题结论）∴CQ：PQ：PC = PA:PB:PC=3:4:5∴满足CQ²+PQ²=PC²∴△PQC是直角三角形