Chapter 1 Proof (2) 1. 1 Can you prove them (1) 1? Trilateral equivalence: two triangles are congruent; 2. Two sides and included angles are equal: two triangles are congruent; 3. Two angles and sides are equivalent: two triangles are congruent; 4. Corresponding angle, corresponding edge; 5. Two triangles with two angles and opposite sides of one angle are congruent; 6.； 7. The bisector of the vertex angle, the midline of the bottom edge and the height of the bottom edge; 8. Equal,; 9.c； 10.c； 1 1.a； 12.c； 13. 17cm； 14.； 15.； 16.； 17. Hint: proof; 18.； When point D is the midpoint of BC, DE=DF (hint: proof). 1. 1 Can you prove them (2) 1. 2. 18 or 21; 3. A triangle with equal height on both sides is an isosceles triangle, true; 4.c； 5.d； 6. isosceles; 7.5cm8.b； 9. Tip: Proof; 10. Hint: use "SSS" to prove it; 1 1. 12. Yes; 13. Hint: proof; Among them:; 14. Hint: B is the extension line of BM perpendicular to FP at point M; Gather sand and make a tower (1) Tip: Proof; (2) acute triangle; (3) ; 1. 1 Can you prove them? (3) 1.( 1) isosceles (2) equilateral (3) equilateral; 2. one, three; 3.a； 4.b； 5.a； 6.4, ,2; 7.8; 8.c； 9.Be = 1 hint: certificate; 10. Omit; 1 1. 12.( 1) ; (2) Prompt by (1). Gather sand into a tower (1): proof; (2) ellipsis; (3) established; 1.2 right triangle (1) 1. 12,10; 2.； 3.5, ; 4. Equiangular is diagonal; 5.3; 6.b； 7.a； 8.d； 9.b； 10.30; 1 1.( 1)60,6 1(2)35,37; 12. Hint: more than d; 13. The area is prompt: link AC; 14. Tip: Find the area of right-angled trapezoid and derive the trilateral relationship of right-angled triangle; 15. Right triangle; Collect sand in a tower for 2 seconds; 1.2 right triangle (2) 1. A set of right angles and hypotenuses, HL; 2.3; 3.HL，，AAS4.d； 5.b； 6.b； 7. prompt: link be8. prompt: certificate; 9. Omission; 10. If the extension line of BA and CE intersects at point F, it can be proved that CE=EF, and then it can be proved that: (ASA); 1 1.( 1) Tip: Certificate first, then certificate; (2) ellipsis; Collect sand into a tower; Perpendicular bisector (1) 1. 1.3 line segment is equal, which is on the vertical line of this line segment; 2.a； 3.5, 10, ; 4. perpendicular bisector; 5. BC; 6.4; 7.c； 8.； 9. Omission; 10.5cm, hint: link advertisement; 1 1.9cm； 12.( 1) omitted; (2)CM = 2BM； 13.a； Tips for gathering sand into a tower: certificate; 1.3 The median vertical line of the line segment (2) 1. Eccentricity and equality; 2. obtuse triangle, acute triangle and right triangle; 3. Equality; 4.； 5.d； 6.4; 7.( 1) A (2) Take the midpoint D of BC, and take point D as the vertical line of BC (3) Tangent point A on the vertical line, so that AD=h (4) AB, AC; 8.( 1) 10 Tip: The circumference of △BCE is BE+EC+BC=25, BE = AE, AC = AE+EC; (2) Prompt: ask ∠ ABC = ∠ C = 72 first, and then ∠ BEC = 72, so ∠ BEC = ∠ C; 9.( 1) 12 (2) (3) equilateral triangle; 10. Hint: certificate; Tip: link am; On the bisector of the angle 1.4 (1) 1. Angular bisector; 2.=; 3.=; 4. 1; 5.b； 6.c； 7.； 8. Omission; 9. Tip: Certificate; 10.( 1) Hint: Do as above N (2); 1 1. 12. Hint: link OA; (1) Proof: Again; ; ; Again,; ; Again; ; ; ; (2) when,; ; ; ; Again; The quadrilateral is a parallelogram; ; 1.4 bisector (2) 1. Inside, three sides of the triangle; 2.( 1)8,(2)8,(3)3; 3.40, 130; 4.c； 5.a； 6. hint: link AO to do it; 7. Omission; 8. Intersection point of angular bisector; 9.( 1) Omit (2); 10. hint: use m to prove; 1 1. Hint: Link DC,; 12. 10cm； Figure (2) Conclusion: FG= (AB+AC-BC) Hint: Extend AG and AF respectively, and BC intersects at point M and point N, then FG= MN ... Figure (3) Conclusion: FG = (AC+BC-AB); Unit comprehensive evaluation 1. b； ; 2.c； 3.b； 4.c； 5D； 6.b； 7.a； 8.c； 9.c； 10.20; 1 1.8; 12.28; 13.； 14. Isosceles; 15. Equality; 16.； 17. Omit; 18. Hint: certificate; 19 . 4 . 5cm； 20. Tip: Certificate; 2 1. Hint: certificate; 22. Tip: Certificate; 23. Tip: Certificate; 24. Tip: Certificate; The second chapter is quadratic equation 2. 1 how wide the lace is 1. c； ; 2.d； 3.b； 4.d； 5.b； 6.4x2- 1=0，4，0，- 1； 7 . a≠ 1； 8.m≠ 1 and m≠3, m =-3; 9.2+ ; 10.5; 1 1.4 ; 12.( 1)k≦，(2)k = 1； 13.30; Sand accumulation into a tower (1) k ≦-1; (2)b≦； 2.2 Matching method (1) 1.5 or-1; 2.0 or 5; 3.c； 4.b； 5.b； 6.c； 7.( 1)x =； (2)x =； (3)x 1=5，x2 =-3； (4)x 1=，x2 =； (5)x 1= - 1+，x2 =- 1-； (6)x 1= -4+3，x2 =-4-3； 8.x 1= - 1，x2 =-2； 9.( 1) The original formula =6(x- 1)2+ 12, no matter what the value of x is, it is 6 (x-1) 2+12 > 0 ; (2) the original formula =- 12(x+ 2-, regardless of the value of x is-12(x+ 2-