∠∠BCE+∠ACE = 90,∠ACE+∠CAE=90,
∴∠BCE=∠CAE.
∵AC⊥BC,BF∥AC.∴BF⊥BC.
∴∠ACD=∠CBF=90,
AC = CB,
∴△ACD≌△CBF.∴CD=BF.
∵CD=BD=? 12BC,∴BF=BD.
∴△BFD is an isosceles right triangle.
∫∠ACB = 90,CA=CB,
∴∠ABC=45。
∫∠FBD = 90 degrees,
∴∠ABF=45。
∴∠ABC=∠ABF, that is, BA is the bisector of∠∠ ∠FBD.
∴BA is the high line of FD edge, and Ba is the middle line of FD edge.
That is, AB divides df vertically.
Model essay for primary school students' third-grade speech 1
Dear teachers and students:
Hello everyone!
Happ