直线与圆相交与点E、F，EF为圆的直径，且AE=EF=FB,直线AP与 圆 de半径OD垂直与D，求证： 角相等

证明: 设BD与圆O交于G,连EG
设圆O半径为R,则,AE=EF=FB=2R
AD⊥OD==>AP为圆O切线==>AD^2=AE*AF=2R*4=8R^2,AD=2√2R
cosA=AD/AO=(2√2R)/(3R)=2√2/3
BD^2=AD^2+AB^2-2AD*AB*cosA
=8R^2+36R^2-2*(2√2R)*(6R)*(2√2/3)=12R^2
BD=2√3R

BG*BD=BF*BE
BG=BF*BE/BD=2R*4R/(2√3R)=4√3/3R
BG/BD=(4√3/3R)/(2√3R)=2/3
BE/BA=(4R)/(6R)=2/3
==>BG/BD=BE/BA==>EG‖AP
==>∠ADE=∠DEG
AP切圆O于D==>∠PDB=∠DEG
==>∠ADE=∠PDB