Question

正方形ABCD中延长AD到E使DE=AD再延长DE到F使DF=BD连接BF交CE于P交CD于Q求证PD=PQ 求辅助线法

Answer 1

在正方形中ABCD ∠ADB=∠DBC=∠BDC=45º DF=BD ∴∠DBF=∠DFB

∠ADB=∠DBF+∠F ===>∠DBF=∠DFB=22.5º

===>∠QBC=45-∠DBF=45-22.5=22.5º

===>∠DQP=∠BQC=90-∠QBC=90-22.5=67.5º

DE=AD=DC DCE=45º

∠EPF=∠BPC=180-∠PBC-∠BCD-∠DCE=180-22.5-90-45=22.5º=∠F ∴EP=EF

∵DF=BD=EC EP=EF ∴PC(EC-EP)=DE(DF-EF)=DC 又∵∠DCP=45º

∴∠QDP=(180-∠DCP)/2=(180-45)/2=67.5º=∠DQP

∴PD=PQ