Question

初二数学题：已知AC=BC,AE=AD,AE⊥AD,EF⊥AC,AC⊥BC,BD/BC=1/2,BP/PE=1,求证AP

Answer 1

证：∵EF⊥AC

∴∠EFP=∠C=90°,∠AFE=90°

∵BP=PE,∠EPF=∠BPC

∴△EPF全等△BPC

∴FP=CP=CF/2

∵AE⊥AD

∴∠CAD+∠EAF=∠DAE=90°

∵∠CAD+∠ADC=90°

∴∠EAF=∠ADC

∵AD=AE,∠AFE=∠C=90°

∴△ADC全等△EAF

∴AF=CD=BC/2

∵AC=BC

∴AF=CF=AC/2,FP=CP=AC/4

∴AP/CP=（AF+FP）/CP=3