Question

如图，△AOB中，OA＝OB,∠AOB＝90°，BD平分∠ABO交OA于D，AE⊥BD于E，求证：BD＝

Answer 1

因AE⊥ BE ，AO⊥ BO ，故∠ EAD=OBD=ABD= 22.5°因BD平分∠ABO ,故 OD=AD*BO/AB =√2 /2 * AD由Rt△AED ,有 AE= AD* cos22.5° ,而在Rt△BOD中 ,BD=OD /sin22.5°=√2/2 AD /sin22.5°=(sin 45°/sin22.5°) * AD =2ADcos22.5° 所以 BD=2AE

Answer 2

延长AE交BO延长线于F
∵AE⊥BE
∴∠AEB=∠FEB=90°
∵BD平分角ABO
∴∠ABE=∠FBE
∵BE=BE
∴△ABE ≌△FBE
∴AE=FE
∴AF=2AE
∵∠AEB=∠AOB=90°
∴∠OAF+∠ADE=90° ∠OBD+∠ODB=90°
∵∠ADE=∠ODB
∴∠OAF=∠OBD
∵OA=OB ∠AOF=∠BOD
∴△AOF≌△BOD
∴AF=BD
∴BD=2AE

Answer 3

很简单，BD=EB-ED