FG⊥AD at point g, FM⊥BC at point m, FN⊥AE at point n,
∵∠ 1=∠2,
∴GF=MF
∫∠3 =∠4
∴MF=FN
∴GF=FN
Point F is on the bisector of ∠DAE.
3. Prove that (1)∵PC⊥OA is at point C and PD⊥OB is at point D,
∴∠OCP=ODP=90
∠∠AOP =∠BOP,OP = OP。
∴△OCP=△ODP(AAS)
∴OC=OD
(2), OC = OD, and OP is the bisector of ∠AOB.
Perpendicular bisector whose op is CD.
∴