1.∠4,∠4,∠2,∠5
2. BC 2 1, 3
3.C
4.∠2 equals ∠3, and ∠3 and ∠5 are complementary. The reason is omitted.
5. The isosceles angles are ∠BFD and ∠DEC, and the ipsilateral internal angles are ∠AFD and ∠AED.
6. There are four pairs. The congruence angles are ∠B and ∠GAD, ∠B and ∠DCF, ∠D and ∠HAB, ∠D and ∠ ECB; The internal angles are ∠B and ∠BCE, ∠B and ∠HAB, ∠D and ∠GAD, ∠D and ∠ DCF; The internal angles on the same side are ∠B and ∠DAB, ∠B and ∠DCB, ∠D and ∠DAB, ∠D and ∠DCB.
1.2( 1)
1.( 1)AB, CD (2)∠3, same angle, two straight lines are parallel.
leave out
3.AB‖CD, the reason is omitted
4. It is known that ∠B, 2 has the same angle and two straight lines are parallel.
5.a and B are parallel, and the reason is omitted.
6.DG‖BF。 The reasons are as follows: DG and BF are the bisectors of ∠ADE and ∠ABC respectively.
∠ADG= 1/2
∠ Ade ∠ABF=005
∠ABC, then ∠ADG=∠ABF, and so on.
When the angles are equal and the two straight lines are parallel, DG‖BF is obtained.
1.2(2)
1.(1) 2,4, offset angles are equal, two straight lines are parallel (2) 1, 3, offset angles are equal, and two straight lines are parallel.
2.D
3.( 1)a‖c, the congruence angle is equal, two straight lines are parallel (2)b‖c, the internal dislocation angle is equal, and the two straight lines are parallel.
(3)a‖b, because the antipodal angles of ∠ 1 and ∠ 2 are ipsilateral internal angles and complementary, so the two straight lines are parallel.
4. Parallel. The reasons are as follows: from BCD =120, CDE = 30, DEC = 90.
So ∠ DEC+∠ ABC = 180, AB ‖ DE (the inner angles on the same side are complementary and the two straight lines are parallel).
5.( 1) 180 ; AD； B.C.
(2)AB and CD are not necessarily parallel. If you add the condition ∠ ACD = 90, or ∠ 1+∠ D = 90.
Can explain AB‖CD, etc.
6.AB‖CD。 ∠ Abd+∠ BDC = 180 7。 Omit.
1.3( 1)
1.D 2。 ∠ 1=70 ,∠2=70 ,∠3= 1 10
3.∠3=∠4. The reasons are as follows: from ∠ 1=∠2, DE ∠ BC (the same angle, two straight lines are parallel),
∴∠ 3 =∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠
4. Vertical meaning; Known; Two straight lines are parallel and the same angle is equal; 30
5.β=44 .∫ab‖CD，∴ α=β
6. (1) ∠ b = ∠ d (2) x =10 is from 2x+ 15=65-3x, so ∠ 1 = 35.
1.3(2)
1.( 1) Two straight lines are parallel and equal to the complementary angle. (2) The two straight lines are parallel and the internal dislocation angles are equal.
2.( 1)× (2)× 3.( 1)DAB (2)BCD
4.∫∠ 1 =∠2 = 100, ∴ m ‖ n (internal dislocation angles are equal and two straight lines are parallel).
∴∠4 =∞∠3 = 120 (two straight lines are parallel with the same included angle).
5. Take Canada as an example.
6.∠APC=∠PAB+∠PCD。 Reason: If AC is connected, ∠ BAC+∠ ACD = 180.
∴ ∠PAB+∠PCD= 180 -∠CAP-∠ACP。
∠ APC = 180-∠ Cap-∠ ACP，∴ APC = ∠ PAB+∠ PCD。
1.4
1.2
2.AB is parallel to CD. The measured length of the line segment BD is about 2cm, so the distance between the two poles is about 0.
Yes 120m.
3. 1? 5cm 4。
5. From m ‖ n, AB ⊥ n and CD ⊥ n, we know that AB = CD ∠ Abe = ∠ CDF = 90.
∫AE‖cf，∴ ∠AEB=∠CFD。 ∴△aeb?△CFD，
∴ AE=CF
6.AB=BC。 The reasons are as follows: If AM ⊥l2 is in M and BN ⊥l3 is in N, then △ ABM.
△BCN, get AB=BC.
Review exercises
1.50 2.( 1)∠4 (2)∠3 (3)∠ 1
3.( 1)∠B, two straight lines are parallel and equal to the complementary angle.
(2)∠5, the internal dislocation angles are equal and the two straight lines are parallel.
(Question 5)
(3)∠BCD, CD, the internal angles on the same side are complementary, and the two straight lines are parallel.
4.( 1)90 (2)60
5.AB‖CD。 Reason: As shown in the figure, it is obtained from ∠ 1+∠ 3 = 180.
∠3=72 =∠2
6. From AB‖DF, ∠ 1 = ∠ D = 1 15. From BC‖DE, ∠ 1+∠ B = 65438.
∴ ∠B=65
7.∠A+∠D= 180，∠C+∠D= 180，∠B=∠D
8. incorrect, sketch.
9. Because ∠ DE ∠ BC = ∠1= ∠ 2, de ∠ BC. So ∠ AED = ∠ C = 70.
10.( 1)DC。 The reason is ∠ ab 'e = ∠ b = 90 = ∠ d.
(2) From B'E‖DC, we get ∠ BEB' = ∠ C = 130.
∴∠aeb′=∠aeb = 0.5
∠BEB′= 65
Chapter 2 Special Triangle
2. 1
1.B
2.3; △ABC、△ABD、△ACD； ∠ADC； ∠DAC，∠C； A.D., DC; Alternating current
3. 15cm, 15cm, 5cm 4. 16 or 17.
(Question 5)
5. As shown in the figure, the answer is not unique. Point C 1, C2 and C3 in the figure can be used.
6.( 1) Omit (2)CF= 1? 5cm
7.AP shares ∠BAC. The reasons are as follows: If AP is the center line, BP=
Personal computer And AB=AC, AP=AP, get △ ABP △ ACP (SSS).
∴ ∠BAP=∠CAP
2.2
1.( 1) 70, 70 (2) 100, 40 2.3, 90, 50 3.
4.∠ B = 40, ∠ C = 40, ∠ Bad = 50, ∠ CAD = 50 5.40 or 70.
6.BD=CE。 Reason: From AB=AC, you get ∠ABC=∠ACB.
∠∠BDC =∠CEB = 90°，BC = CB，
∴△BDC?△CEB(AAS)。 ∴ BD=CE
(This problem can also be solved by area method.)
2.3
1.70, isosceles 2.3 3.70 or 40.
4.△BD and CD are isosceles triangles for the following reasons: BD and CD are ∠ ABC and ∠ ACB respectively.
Reference answer
5 1
Branch, get ∠DBC=∠DCB. Then DB=DC.
5.∠DBE=∠DEB，DE=DB=5
6. Both 6.△ DBF and △EFC are isosceles triangles for the following reasons:
∫△ade and △FDE coincide, ∴ △ ade = ∠ FDE.
∫de BC，∴ ∠ADE=∠B，∠FDE=∠DFB，
∴ ∠B=∠DFB。 ∴ DB=DF, that is, △DBF is an isosceles triangle.
Similarly, it can be seen that △EFC is an isosceles triangle.
7.( 1) divide 120 into 20 and 100 (2) divide 60 into 20 and 40.
2.4
1.( 1)3 (2)5
2.△ADE is an equilateral triangle for the following reasons: △ ABC is an equilateral triangle,
∴ ∠A=∠B=∠C=60。 ∫ BC, ∴ ∠ADE=∠B=60,
∠ AED =∠ C = 60, that is ∠ Ade =∠ AED =∠ A = 60.
leave out
( 1) AB ‖ CD。 Because < BAC = < ACD = 60.
(2)AC⊥BD. Because AB=AD, ∠BAC=∠DAC.
5. It is concluded from AP=PQ=AQ that △APQ is an equilateral triangle, then ∠ APQ = 60, BP=
AP，∴∠ B = ∠ BAP = 30。 Similarly, ∠ C = ∠ QAC = 30.
∴ ∠BAC= 120
6.△DEF is an equilateral triangle. The reasons are as follows: ∠ Abe +∠ FCB = ∠ ABC = 60.
∠ABE=∠BCF，∠ FBC+∠ BCF = 60。 ∴∠ DFE = 60。 This is also the case.
∠ EDF = 60, ∴△ def is an equilateral triangle.
7. The answer is not unique, as shown in the figure.
(Question 7)
2.5( 1)
1.C 2.45，45，6 3.5
4. ∵∠ B+∠ C = 90, ∴△ ABC is a right triangle.
5. It can be found that ∠ C = 72, ∠ DBC = 18.
6.DE⊥DF,DE=DF. The reasons are as follows: △ CED △ CFD,
∴ Germany = df. ∠ Equivalent cyclic density = 45, ∴∠ Equivalent cyclic density = 45. Similarly, CDF = 45,
∴∠ EDF = 90, that is, DE⊥DF.
2.5(2)
1.D 2.33 3。 ∠A=65，∠B=25 4。 DE=DF=3m
5.BE=0.5
AC，DE=0.5
AC，get be = de6. 135m。
2.6( 1)
1.( 1) 5 (2) 12 (3) 5 2.A=225
3. Make a right triangle with right angles of 1cm and 2cm respectively, and its hypotenuse is 5cm long.
4.22cm (or 8cm) 5.169cm2
6.18m
7.s trapezoid BCC'D'=0.5
(C′D′+BC)？ BD′= 0.5(a+b)2，
S trapezoid bcc' d' = s △ AC' d'+s △ ACC'+s △ ABC = ab+0.5c2.
From 1/2(a+b)2=ab+0.5c2, a2+b2=c2 is obtained.
2.6(2)
1.( 1) cannot (2) cannot (2) be a right triangle, because m2=p2+n2 is satisfied.
meet
4.∠BAC∠ADB∠ADC are all right angles.
5. If BD is connected, ∠ ADB = 45, BD 槡 = 32.∴ BD2+CD2=BC2,
∴ ∠BDC=90。 ∴ ∠ADC= 135
6.( 1)n2- 1，2n，n2+ 1
(2) It is a right triangle because (N2-1) 2+(2n) 2 = (N2+1) 2.
2.7
1.BC=EF or AC=DF or ∠A=∠D or ∠ b = ∠ e 2.
3. Congruence, based on "HL"
4. From △ Abe △ EDC, AE = EC, ∠ AEB+∠ DEC = 90.
∴∠AEC = 90°, that is, △AEC is an isosceles right triangle.
5.∫∠ADB =∠BCA = Rt∞，AB=AB，AC=BD，
∴ Rt△ABD≌Rt△BAC(HL)。 ∴ ∠CAB=∠DBA，
∴ OA=OB
6.DF⊥BC. The reasons are as follows: Rt△BCE≌Rt△DAE is known,
∴ ∠B=∠D, so ∠ D+∠ C = ∠ B+∠ C = 90.
Review exercises
1.A 2。 D 3.22 4. 13 or119 5.b6. isosceles.
7.72, 72, 48, 7 9.64.
10.ad = ae，∴ ∠ADE=∠AED，∴ ∠ADB=∠AEC。
bd = ec，∴△Abd?△ace。 ∴ AB=AC。
1 1.4? 8 12.B
13. Link BC. ab = ac，∴ ∠ABC=∠ACB。
∠∠Abd =∠ACD，∴ ∠DBC=∠DCB。 ∴ BD=CD。
14.25π
15. If BC is connected, Rt△ABC≌Rt△DCB, ∴ ∠ACB=∠DBC, so OB=OC.
16.AB= 10cm。 ∠AED =∠C = Rt∞，AE=AC=6cm，DE=CD。
BE=4cm。 In Rt△ bed, 42+CD2=(8-CD)2, CD=3cm.