Question

如图,菱形ABCD中,∠A=60°,点E,F分别在边AB,AD上,AE=DF,BF与DE相交于点G

Answer 1

(1)连接BD

∵四边形ABCD是菱形,∴AB=AD

∵∠A=60°,∴BD=AD,∠ADB=60°

∵AE=DF,∴△ADE≌△DBF

∴∠AED=∠DFB

∴AEPF四点共圆.∴∠DFP=60°

(2)延长PB至Q,使BQ=DP,连接CQ

∵∠DFP=∠BCD=60°,∴BCDP四点共圆

∴∠CDP=∠CBQ

∵CD=CB,DP=BQ,∴△CDP≌△CBQ

∴CP=CQ

∵∠CPB=∠CDB=∠DBA=60°,∴CP=PQ

即CP=DP+PB