Current location - Training Enrollment Network - Mathematics courses - 20 14 Page 44 of the first volume of the second grade mathematics textbook 4.5. 1 1. 12 Answer.
20 14 Page 44 of the first volume of the second grade mathematics textbook 4.5. 1 1. 12 Answer.
4. Proof: ∫∠3 =∠4, ∴∠ABD=∠ABC (complementary angles of equal angles are equal).

In ABD and ABC:

∠ 1=∠2,

AB=AB,

∠ABD=∠ABC,

∴δabd≌δab(asa),

∴AD=AC。

5. Prove: In Δ Δ ABC and Δ Δ CDA:

∠ 1=∠2,

∠B=∠D,

AC=CA,

δABC?δCDA(AAS),

∴AB=CD。

1 1. Proof: ∫ab∨ed, ∴∠ B = ∠ E.

∵AC∥DF,∴∠ACB=∠DFE,

BF = CE,

∴BF+EF=CF+E,

That is BC=EF,

In ABC and DEF:

∠B=∠E,BC=EF,∠ACB=∠DFE,

∴δabc≌δdef(aas),

∴ab=de,ac=df>;

12、AE=CE .

Proof: ∫fc∨ab, ∴∠A=∠ECF, ∠ ADE = ∠ F.

In EAD and δECF,

∠A=∠ECF,∠ADE=∠F,DE=EF,

∴δead≌δecf,

∴AE=CE。