Question

如图,已知PA与⊙O相切于A,过O割线PBC交⊙O于B、C,∠APC平分线分别交AB,AC于D,E,证明∠ADE=∠AE

Answer 1

连AO则AO⊥AP于A,AO=CO

∴∠OCA=∠OAC

∵BC为⊙O的直径

∴∠CAO+∠OAB=90°

即∠OCA+∠OAB=90°

∵∠BAP+∠OAB=90°

∴∠OCA=∠BAP①

又∵PE平分∠APC

∴∠APE=∠CPE②

∵∠ADE=∠BAP+∠APE③

∠AED=∠OCA+∠CPE④

由①②③④得

∠ADE=∠AED