Reference answer

First, multiple-choice questions:

1~5 DDBCA

6~ 10 BCDBC

Second, fill in the blanks:

1 1、 ③

12、4

13、 1 10

14,25, and the corresponding sides of congruent triangles are equal.

15, AE=CF, and two sets of corresponding edges of two triangles and their included angles are equal, then two triangles are congruent (that is, edge theorem).

16、70

17、 1.5

18、2

19、3

20、(0,-3)、(-2,-3)

Third, answer questions:

2 1. Make the bisector of the intersection of the railway and the highway, and the intersection with the river is the position of the pier Q. ..

22、

Proof: ∫AB//CD

∴∠BAC=∠ECD

∠∠B =∠D，AC=CD。

∴△BAD≌△ECD(AAS theorem)

∴BC=ED

23、

(1) Proof: ∴∵ad⊥bc fdb =∠CDA = 90.

BD = AD，FD=CD。

So △ BDF △ ADC (SAS theorem)

∴∠FBD=∠CAD

(2) Proof: ∫∠FBD =∠CAD

And ≈BFD =≈AFE

∠ FDB = 180-∠ FBD-∠ BFD。

Finite element analysis = 180-∠ computer aided design -∠AFE

∴∠FEA=∠FDB=90

∴BE⊥AC

24、

Proof: df⊥ac. ∵de⊥ab

∴∠DEB=∠DFC=90

And ∵BE=CF and BD=CD.

∴△DEB≌△DFC(hl theorem)

∴DE=DF

∠∠DFA =∠Deb = 90。

AD=AD

∴△ADE≌ADF(hl theorem)

So ∠EAD=∠FAD

Ad Split ∠BAC

(2)AB+AC=2AE

25、

( 1)AD=DE

(2) Prove that point D is the midpoint of BC.

∴BD=CD

∠∠BDE =∠CDA，AD=DE。

∴△ACD≌△EBD(SAS theorem)

(3) 1

26、

(1) Proof:

∵ABCD is a square.

∴AD//BG ∴∠EAD=∠FGB

It's ∵BF⊥AG again.

∴∠BFG=90，

∴∠FBG+∠FGB= 180 -90 =90

∠∠FBG+∠FBA =∠ABG = 90。

∴∠FBG=∠FBA

∴∠EAD=∠FBA

And ∵DE⊥AG, ∴∠ DEA = ∠ AFB = 90.

AB = DA

∴△ABF≌△DAE(AAS theorem)

(2)BF+EF=AF

(3)①△ABF?△DAE，AF+EF=BF

②EF=AF+BF

EF+AF=BF