Question

如图,圆内接四边形ABCD的外角角DCH等于角DCA,DP垂直于AC ,垂足为P,DH垂直于H,求

Answer 1

证明：连接BD∵DP⊥AC,DH⊥BH∴∠DPC=∠DHC=90°又∵∠DCH=∠DCA,DC=DC∴△DCH≌△DCP（AAS）∴CH=CP【1证毕】 DH=DP又∵∠DHB=∠DPA=90° ∠DBH=∠DAP（同弧所对的圆周角相等）∴△DBH≌△DAP（AAS）∴AP=BH