Question

如图16,AC与BD相交于点O,BE、CE分别平分∠ABD、∠ACD,且BE与AC交与F点,CE与BD相交于P点.∠A=

Answer 1

∠A=60°, ∠D=50°, ∠AOB=∠DOC, 故∠ABO+10°=∠DCO

∠OFE=∠AOB+∠FBO=∠AOB+∠ABO/2

∠OPE=∠DOC+∠PCO=∠AOB+∠ABO/2+5°

∠FOP=∠A+∠ABO=∠ABO+60°

∠AOB+∠ABO=180°-∠A=120°

∠E=360°-∠OFE-∠OPE-∠FOP

=360°-65°-2(∠AOB+∠ABO)

=360°-65°-240°

=55°