3. Geometry and reasoning.
1. Connect AC, because AD=4 and CD=3, so it is obtained from Pythagorean theorem: AC=5, and because AB = 13 = 169, BC+AC = 12+5 = 65438.
=( 12×5)/2-(3×4)/2=30-6=24 .
2.AB，AC
3. Because AB=AC, ∠ ABC = ∠ C. Because BC=BD, ∠C=∠BDC=∠ABC. Because AD=DE=BE, ∠ A = ∠ AED = 2.
4.( 1)∫DE is the center line of △ABC. ∴ DE = 1/2ac，CE = be。 ∫CF = 1/2ac。 ∴。 CF =
(2)∫≈ 1 =∠2∴BM = am。 ∵ AB//CD。 ∴∠ 1 = ∠ CMB，∠2 =∞。
③∫e is the midpoint of BC, ∴BE=EC is ∵EF⊥AB, F,EG⊥CD is G, and EF=EG, ∴△BEF≌△EGC(HL), ∴.
(4) If the extension line of the intersection D from DE‖AC to BC is at E, then ∠ BDE = 90. Then d stands for DF⊥BC at F.
∵ trapezoid ABCD is isosceles trapezoid, then AC=BD. And \ad \bc, DE \ac. A quadrilateral is a parallelogram. AD = CE，AC=DE=BD。
∴△BDE is an isosceles right triangle. Then df =1/2be =1/2 (BC+ce) =1/2 (BC+ad) =1/2 (m+n) ∴.
(5) In the isosceles trapezoid ABCD, AB=CD ∠ABC=∠DCB BC is the public side, ∴△ABC≌△DCB(SAS).
∴AC=BD advertising is the male side of∴△ Abd ≌△ DCA (SSS).
∴∠abo=∠dco∠AOB =∠doc△ABO?△dco(AAS)
∴ OB = OC ∠ BOC = 60 ∴△ OBC is an equilateral triangle. Similarly, △AOD is an equilateral triangle.
Even DE and CF, from ef =1/2ab =1/2cd (ef is the midline of △AOB) ∴EF=DG=CG(G is the midpoint of CD).
E is the midpoint of AO. ∴DE is the middle vertical line of AO. ∠ DEC = 90。 EG is the center line of ∴ ef = DG = eg. ∴ DG = eg.
CF is also the middle vertical line of BO. ∴∠ DFC = 90 ∴ FG = CG = EF ∴△ EFG where EF=EG=GF. △ EFG is an equilateral triangle. 5( 1)∫ABCD∴ad//DC∶AC//MQ。 AM//CQ ∴ Quadrilateral ACQM is parallelogram: AC//PN, and AP//CN ∴ Quadrilateral ACNP is parallelogram: PN.
∵ parallelogram acnp ∴ AC = pn ∴ MQ = pn ∵ MP = MQ-pq, qn = pn-pq ∴ MP = qn.
(2)∫AE = cf, AF = CF-AC, ce = AE-af∴af = ce∫ parallelogram ABCD∴AD=BC,AD//BC.
∴∠dac=∠acb∠DAC+∠daf = 180，∠ACB+∠BCE= 180 ∴∠DAF=∠BCE
In △DAF and △BCE, DA = BC, ∠ DAF = ∠ BCE, AF = CE ∴△ DAF △ BCE (SAS).
∴ df//be ∵ df = be, df//be, ∴ df = be, ∠ DEA = ∠ CEB quadrilateral BEDF is a parallelogram.
(3)∫ parallelogram ABCD ∴ DC//AB, DC = ab∴∠ECN =∠fam∫DC = ab, DE=BF and
EC= DC-DE, FA=AB-BF ∴EC=FA in △ECN and △FAM, EC=AF, ∠ECN=∠FAM,
cn=am∴△ecn≌△fam(sas)∴en=fm,∠enc=∠fma,∵∠enc+∠enm = 180，
∠ fma+∠ FMN =180 ∴ EMN = ∠ FMN ∴ en//MF: en = FM, en//FM, ∴ quadrilateral MFNE is a parallelogram.
(4)1:parallelogram ABCD∴AD//BC ∴AE//CF ∴∠AEO=∠CFO. In △AEO and △CFO,
∠ AEO =∠ CFO，∠ AOE =∠ COF，ao = co∴△AEO?△CFO(AAS)∴AE = CF∶AE//cf，AE = cf，∴.
2)ef⊥AC∴eoa =∠EOC = 90 in△eoa and △EOC, EO = EO, ∠ EOA = ∠ EOC, OA = oc∴△eoa∞.
3)EF=AC,EF⊥AC
6.( 1) Let y=kx. When y=24 and x=6, 24=6k and the solution k=4 ∴y=4x.
Let a set of equations of y=kx+b be 24= 14k+b, the solution of 0=20k+b is 5=-4, and b=80 ∴y=-4x+80.
(2) When x=4, Y = 4x = 4x4 = 16 When x= 18, Y =-4x = 80 =-4x 18+80 = 8.
(3) In y=4x, when y=20, 20=4x gives x=5 x on the AB side.
In y=-4x+80, when y=20, 20=-4x+80 gives x= 15 x on the DC side.
Query operation
1 and (1) quadrilateral EFGH are parallelograms.
H.G is the midpoint of AD. CD ∴HG is the bit line in △DAC ∴hg= 1/2ac Hg//AC.
E.H is the midpoint of AB. AD ∴EH is the center line of △ABD ∴EH= 1/2BD, EH//BD.
E.F is the midpoint of AB. CB ∴EF is the center line of △BCA ∴EF= 1/2AC, EF//AC.
F.G is the midpoint of CB. CD ∴FG is the center line of △ CDB ∴FG= 1/2BD,FG//BD.
∵ Hg//AC, ef//AC ∴ Hg//ef: eh//BD, gf//BD ∴ eh//gf: Hg//ef, EH//GF∴ quadrilateral EFGH is a parallelogram.
(2) When the diagonal of quadrilateral ABCD is vertical, quadrilateral EFGH is a rectangle (intersecting HG. BD is the intersection of j and AC. BD is the intersection of I and GF. AC is k).
∵DB⊥AC, ∴∠ DIC = 90 ∵ HF//AC, GF//db ∴ quadrilateral JIKG is a parallelogram ≈DIC = 90.
∴ parallelogram JIKG is a rectangle∴∠ HGK = 90∴∴ parallelogram EFGH is a rectangle.
(3) When the diagonals of quadrilateral ABCD are equal, quadrilateral EFGH is rhombic.
* Hg = 1/2ac，ef = 1/2ac，eh = 1/2bd，fg = 1/2bd，AC=BD∴HG=EF=EH=FG∴.
(4) When the diagonals of quadrilateral ABCD are vertical and equal, quadrilateral EFGH is a square.
2.( 1)∵ parallelogram ABCD∴AD//BC∴∠DAF=∠AFB∵AF bisection ∠BAD∴∠BAF=∠FAD.
∴∠BAF =∠AFB∴ab = BF = 3cm∴cf = BC-BF = 2cm÷ad//BC∴∠ade =∠dec÷ed。
∠cda∴∠cde=∠eda∴∠ced=∠cde∴cd=ec=3cm∴be=cb-ce=2cm
∵ BC = be+ef+fc ∴ ef = BC-be-cf =1cm, that is, BE=2cm, EF = 1cm and CF = 2cm.
(2) CE = 3cm, BF = 3cm, and E.F conforms to ∴BC=EC+BF=6cm.
(3) Omission
3、( 1) 1000 800(2)30 (3)50
physics
1-5 CBCDC 6- 10 ddcab 1 1- 15 bbcac 16-20 caacb 2 1-23 ACA
24. Vibration, air 25. Reduce noise. Scaling ratio, 5.50.
27. Telescopes, microscopes, thermometers hitting the glass wall obliquely, the moon, clouds and the theory of relativity.
30. Constant, endothermic, ascending, endothermic 3 1, swift, 0.5 32, evaporation, inhalation, liquefaction 33, variable speed, simultaneous passing distance 34, loudness, tone 35, stethoscope, sound control lamp 36, solidification, liquefaction, sublimation and melting.
37. optical center, optical center, image, u > 2f, inversion, reduction, f < u < 2f, inversion, magnification, solid, u < f, upright,
Magnification, imaginary number 38, (1)(3), (2), (4)(5), convex lens, inverted, slightly larger than 40-47.
48, sound has energy, sound can spread in the air 49, liquefaction, water vapor, constant 50, slightly 5 1, BACD.
52. Both sides are equal, in the same plane 53, vertical, which is convenient for measuring the object distance and image distance. Replace the chess piece B with a light screen, and the image on the light screen is not empty.