關(guān)于畢達哥拉斯定理證明的論文

1、關(guān)于畢達哥拉斯定理的證明專業(yè)：×××××姓名：××指導老師：××摘要：對于幾何原本中畢達哥拉斯定理的證明過程，歐幾里得以定義，公設，公理的方式進行推理，現(xiàn)將所有涉及畢達哥拉斯定理的證明命題提出。關(guān)鍵詞：畢達哥拉斯定理，定義，公設，公理。正文：定義：1. 點是沒有大小的東西2.線只有長度而沒有寬帶3.一線的兩端是點4.直線是它上面的點一樣地平放著的線5.面只有長度和寬帶6.面的邊緣是線7.平面是它上面的線一樣地平放著 8. 平面角是在一平面內(nèi)但不在一條直線上的兩條相交線相互的傾斜度.9. 當包含角的兩條線都

2、是直線時，這個角叫做直線角.10. 當一條直線和另一條直線交成鄰角彼此相等時，這些角每一個被叫做直角，而且稱這一條直線垂直于另一條直線。11. 大于直角的角稱為鈍角。12. 小于直角的角稱為銳角13. 邊界是物體的邊緣14. 圖形是一個邊界或者幾個邊界所圍成的15. 圓：由一條線包圍著的平面圖形，其內(nèi)有一點與這條線上任何一個點所連成的線段都相等。16. 這個點（指定義15中提到的那個點）叫做圓心。17. 圓的直徑是任意一條經(jīng)過圓心的直線在兩個方向被圓截得的線段，且把圓二等分。18.半圓是直徑與被它切割的圓弧所圍成的圖形，半圓的圓心與原圓心相同。19.直線形是由直線圍成的.三邊形是由三條直線圍成

3、的,四邊形是由四條直線圍成的,多邊形是由四條以上直線圍成的.20.在三邊形中,三條邊相等的,叫做等邊三角形;只有兩條邊相等的,叫做等腰三角形;各邊不等的,叫做不等邊三角形.21.此外,在三邊形中,有一個角是直角的,叫做直角三角形;有一個角是鈍角的,叫做鈍角三角形;各邊不等的,叫做不等邊三角形.22.在四邊形中,四邊相等且四個角是直角的,叫做正方形;角是直角,但四邊不全相等的,叫做長方形;四邊相等,但角不是直角的,叫做菱形;對角相等且對邊相等,但邊不全相等且角不是直角的,叫做斜方形;其余的四邊形叫做不規(guī)則四邊形.23.平行直線是在同一個平面內(nèi)向兩端無限延長不能相交的直線.0公理：1.等于同量的彼

4、此相等2.等量加等量，其和相等；3.等量減等量，其差相等4.彼此能重合的物體是全等的5.整體大于部分。公設: 1.過兩點能作且只能作一直線；2.線段(有限直線)可以無限地延長；3.以任一點為圓心,任意長為半徑,可作一圓；4.凡是直角都相等；   5.同平面內(nèi)一條直線和另外兩條直線相交，若在直線同側(cè)的兩個內(nèi)角之和小于180°，則這兩條直線經(jīng)無限延長后在這一側(cè)一定相交。作圖證明：1.在一個已知有限直線上作一個等邊三角形設AB是已知直線以A為圓心，以AB為距離畫圓以B為圓心，以AB為距離畫圓兩圓交點C到A,B的來連線CA,CBAC=ABBC=BACA=CB=A

5、BABC是等邊三角形2.過直線外一已知點作一直線平行于已知直線。設A是已知點，BC是已知直線，要求經(jīng)過A點做直線平行于BC在BC上任取一點D，連接AD，在直線DA上的點A，做 DAE=ADC設直線AF是直線EA的延長線直線AD和兩條直線BC,EF相交成彼此相等的內(nèi)錯角EAD，ADC EAFBC作畢3.在已知線段上作一個正方形。設AB是已知線段，要求在線段AB上作一個正方形令AC是從線段AB上的點A所畫的直線，它與AB成直角取AD=AB過點D做DE平行于AB，過點B做BE 平行于AD，所以ADEB是平行四邊形AB=DE,AD=BE又AD=AB平行四邊形ADEB是等邊的BAD+ADE=180

6、76;BAD 是直角ADE是直角平行四邊形中對邊及對角相等ABDE是正方形 4：由已知直線上一已知點做直線與已知直線成直角解：設在AC上任意取一點D，使CE=CD在DE上作一個等邊三角形FDE連接FCDC=CECF=CFDF=CFDF=FEDCF=ECF他們是鄰角，由定義10，二者都是直角作畢。5：已知兩條不相等的線段，試由大的上邊截取一條線段是它等于另外一條設AB，C是兩條不相等的線段由A取AD等于線段C以A為圓心，AD為距離畫圓DEFA是圓DEF的圓心AE=AD又C=ADAE=C=AD作畢命題證明：命題1：如果兩個三角形有兩邊分別等于兩邊，而且這些相等的線段所夾的角相等。那么，它們的底邊等

7、于底邊，三角形全等于三角形，而且其它的角等于其它的角，即那等邊所對的角。證明：設ABC,DEF是兩個三角形，AB=DE,AC=DF， BAC=EDF如果移動三角形ABC到DEF上，若A落在點D上，且線段落在DE上AB=DEB與E重合又AB與DE重合 BAC=EDFAC與DF重合又AC=DFC與F重合ABC與DEF重合，即全等命題2：一條直線和另一條直線所交成的角，或者是兩個直角，或者是它們的和等于2個直角證明：設任意直線AB交CD成角CBA,ABD若CBA=ABD則CBA=ABD=90°（定義10）若二者不是直角作BECD于BCBE=EBD=90°CBE=CBA+ABECB

8、E+EBD=CBA+ABE+EBD同理，DBA+ABC=DBE+EBA+ABCCBE+EBD=DBA+ABC=180°原命題得證命題3：對頂角相等證明：設直線AB,CD相交于點EDEA+CEA=CEA+BEC=180°（命題2）DEA=BEC命題4：兩直線平行，同位角相等設直線EF與兩條平行直線AB,CD相交假設AGH不等于GHD不妨設AGH較大AGH+BGH>GHD+BGH又AGH+BGH=180°（命題1）GHD+BGH<180°二直線延長一定會相交又兩直線平行AGH=GHD又AGH=EGB（命題3）GHD=EGB原命題得證命題5：如果在

9、兩個三角形中，一個的兩個角分別等于另一個的兩個角，而且一邊等于另一個的一邊，即過著這邊是的等角的家變，或者是等角的對邊，則它們的其他的邊也等于其他的邊，且其他的角也等于其他的角證明：如果ABDE不妨設ABDE取BG等于DE連接GCBG=DEBC=EFGB=DEBC=EFGBC=DEFGC=DF又GBCDEF其余角和邊也相等（命題1）GCB=DFEBCG=BCA這是不可能的AB=DE又BC=EFAB=DEBC=EFABC=DEF AC=DFBAC=EDF（命題1）假設BCEF不妨設BCEF令BH=EF連接AHBH=EFAB=DE所成的夾角相等AH=DFABHDEFBHA=EFD又EFD=BCA因

10、此，在三角形AHC中，外角BHA等于BCA這是不可能的BC=EF又AB=DE夾角也相等（命題1）ABCDEFAC=DF命題6：在平行四邊形中，對邊相等且對角線二等分其面積（注：幾何原本原文中無平行四邊形的定義定義: 在同一平面內(nèi)兩組對邊分別平行的四邊形叫做平行四邊形。（1）如果一個四邊形是平行四邊形，那么這個四邊形的兩組對邊分別相等。（2）如果一個四邊形是平行四邊形，那么這個四邊形的兩組對角分別相等。）證明：ABCDABC=BCDACBDACB=CBD（命題4）又BC=BCABCDCBABC=BCD又CBD=ACBAC=ACABDACDBAC=CDB平行四邊形ABCD中，對邊對角彼此相等（1）

11、（2）性質(zhì)得證）同樣地，ABCDCB對角線BC平分平行四邊形ACBD的面積命題7：在同底且在相同兩平行線之間的平行四邊形面積相等證明：設ABCD，EBCF是平行四邊形，它們在同底BC。且在相同的平行線AF，BC之間ABCD是平行四邊形AD=BC同理，EF=BC,AD=EFAE=DF又AB=DCFDC=EABEABFDCEB=FC面積EAB-DGE=FDC-DGE面積ABGD=EGCF同加上GBC平行四邊形ABCD面積等于平行四邊形EBCF命題8：如果過任意一條直線上一點有兩條直線不在這一直線的同側(cè)，且和直線所成鄰角和等于二直角，則這兩條直線在同一條直線上證明：如果BD與BC不共線假設BE和CB

12、共線AB在直線CBE之上ABC+ABE=180°（命題2）又ABC+ABD=180°CBA+ABE=CBA+ABD兩邊同時減去CBA則ABE=ABD（公設4，公理1，公理3）這是不可能的BE，BC不共線同理除BD外沒有其他直線與BC共線CB與BD共線命題9：在同底上且在相同兩平行線之間的三角形面積相等證明：如圖所示，設三角形ABC,DBC同底且在相同兩平行線AD，BC之間延長AD和DA分別至F，E，過B作BE平行于CA，過C作CF平行于BD則四邊形EBCA和DBCF都是平行四邊形，且面積相等（命題5）ABC的面積是偶像是必須EBCA的一半 DBC的面積是平行四邊形DBCF的

13、一半（命題6）DBC面積等于ABC的面積命題10：如果一個平行四邊形和一個三角形既通敵又在兩平行線之間，則平行四邊形的面積是三角形的2倍證明：連接ACABC與EBC又同底BC，又在平行線BC和AE之間ABC的面積等于EBCAC平分平行四邊形ABCD平行四邊形ABCD的面積是EBC的2倍平行四邊形ABCD的面積是EBC的2倍關(guān)于畢達哥拉斯定理的證明：直角三角形的直角邊的平方和等于斜邊的平方。已知：如圖所示，ABC是直角三角形。求證：AB²+AC²=BC²。證明：分別以直角邊AB,AC和斜邊BC的作正方形ABFG，正方形ACKH,正方形BCED；（作圖3） 過A作AL

14、平行于BD或CE，連接AD，F(xiàn)C； BAC=BAG=90° C,A,G共線（命題8） 同理，B,A,H共線 DBC=FBA 所以DBC+ABC=FBA+ABC 即DBA=FBC（公理2） 又DB=BC FB=BA 所以ABDFBC（命題1） 平行線AL與BD之間 平行四邊形BL的面積是ABD的2倍 同理，正方形GB的面積是FBC的2倍 由公理2，平行四邊形BL的面積與正方形BD相等（命題10） 同理可得，平行四邊形CL等于正方形HC 正方形BCED的面積等于正方形ABFG與正方形ACKH面積之和（公理2） BC²=AB²+AC² 原命題得證參考文獻：歐幾

15、里得幾何原本The proof of the Pythagorean theorem about Professional: ××Name:××Teacher: ××Abstract: for the geometry of the proof of the Pythagorean theorem was process, to define the kansai, axioms, justice way reasoning, now will all concerned proof of the Pythagorean theore

16、m put forward proposition. Key words: the Pythagorean theorem, definition, axioms, justice. Text:Definition: 1. The point is not part of the things 2. Line length and not only broadband 3. A at both ends of the line is the point 4. Straight line is on it to the point of being the same line 5. Faces

17、only length and broadband 6. The edge is line 7. The plane is on it as a lie flat line 8, is in a plane within intersects each other but not in a straight line of the two intersecting line the gradient of each other. 9. When including Angle of two lines are straight line, the horn is called straight

18、 line Angle. 10. When a straight line and the other hand in a straight line into LinJiao equal to each other, and these horns every called right Angle, and says that a straight line perpendicular to the other in a straight line. 11. Greater than the horns of the right Angle called obtuse Angle. 12.

19、Less than the right Angle called acute Angle 13. The boundary is the edge of the object 14. The figure is a boundary or surrounded by several boundary 15. Round: by a line of surrounded by the plane figure, it is a little and the line any point joined the line are equal. 16. The point (refers to the

20、 definition of the points mentioned in 15) called circle. 17. Circle diameter is any a circular straight after the two direction was round intercepts line, and the round two parts. 18. Semicircle is diameter and was it the circular arc of the cutting that surrounded the graphics, semicircle circle a

21、nd the same circle. 19. Linear form is surrounded by line. Trilateral form by three straight line is surrounded, quadrilateral by four straight lines is surrounded, polygons by more than four straight line is surrounded. 20. In the shape of 3, 3 sides equal, called an equilateral triangle; Only two

22、edges equal, called an isosceles triangle; The edge of the range, called not an equilateral triangle. 21. In addition, in the shape of the trilateral, have a right Angle is, is called a right triangle; Have a Angle is the nails, the nails called triangle; The edge of the range, called not an equilat

23、eral triangle. 22. In the quadrilateral, tote is equal and four Angle is the Angle, is called a square; Angle is a right Angle, but quadrilateral not all equal, called the rectangle; Four equal, but not the right Angle, called diamond; Diagonal is equal and opposite sides equal, but not all equal an

24、d edge horn is not the right Angle, called the inclined square; The rest of the quadrilateral called irregular quadrilateral. 23. Parallel lines are in the same plane introverted ends extend unlimited cannot at the intersection of straight line. 0 Justice: 1. Equal to about the same amount of equal

25、to each other 2. Add amount equal, its and equal; 3. Reduced amount equal, the poor are equal 4. Each other can overlap object is congruent 5. The whole is greater than the partially. Axiom: 1. A can only be made two and a straight line; 2. The line (limited linear) can be infinite extension; 3. As

26、a little to the right to, any long for radius, can make a circle; 4. All right Angle are equal; 5. With plane within a straight line and another two straight line intersection, if in line with the side of the sum of the two an internal Angle is less than 180 °, then these two straight lines aft

27、er the infinite extension in the side must intersect. Drawing the proof: 1. In a given limited on a straight line equilateral triangle Set AB is known straight line With A to the right, to draw circles AB distance With B to the right, to draw circles AB distance Two round) to A C, B to attachment of

28、 CA, CB AC = AB BC = BA CA = CB = AB enables delta ABC is an equilateral triangle 2. A known point for a straight line parallel to the known straight line. Set A is known point, BC is known straight line, after A request to do A straight line parallel to BC Take A little D took office in BC, connect

29、ion AD in straight DA points on A, do < DAE = < ADC A straight line is straight line EA AF linear AD and two straight lines BC, EF into each other NaCuoJiao intersection equal EAD, ADC EAF BC 3. In line for a known on the square. Line AB is a known, in the line AB requirements on a square The

30、line AB to AC from point A are painting of the straight line, it and AB, at right angles Take AD = AB Lead point D do DE, parallel to the AB, lead point B do BE parallel to the AD, so ADEB is a parallelogram AB = DE, AD = BE And AD = AB parallelogram ADEB is equal sides < BAD + < ADE = 180

31、76; < BAD is right angles < ADE is right angles parallelogram edge and diagonal in equal ABDE is a square 4: known line by a known to do a straight line and linear known at right angles Solution: take a little arbitrary in AC D, make CE = CD In DE make one FDE equilateral triangle Connection F

32、C DC CE CF = CF DF = CF DF = FE < DCF = < ECF They are LinJiao, by definition 10, both is right angles Proposition proof:Proposition 1: if two triangle has both sides were equal to both sides, and the equal line between equal the Angle. So, they are equal to the lower side of the bottom edge,

33、triangle is equal to the triangle, and other Angle is equal to other Angle, namely that the Angle to the sides. Proof: set ABC, DEF is two triangles, AB = DE, AC = DF, < BAC = < EDF If mobile triangle ABC to DEF, if A fall in point D, and line in the paragraph DE AB = DE B and E coincidence An

34、d AB and DE superposition < BAC = < EDF AC and DF superposition And AC = DF C and F coincidence enables delta ABC and train DEF coincidence, that is congruent Proposition 2: a straight line and the other a straight line pay into horn, or two right angles, or is their and equal to two right ang

35、les Proof: set any straight line AB/CD into Angle CBA, ABD If < CBA = < ABD The < CBA = < ABD = 90 ° (definition 10) If both not right angles BE as an CD in B < CBE = < EBD = 90 ° < CBE = < CBA + < ABE < CBE + EBD < = < CBA + < ABE + < EBD Similarl

36、y, < DBA + < ABC = < DBE + < EBA + < ABC < CBE + EBD < = < DBA + < ABC = 180 ° Original proposition find Proposition 3: vertical angles equal Proof: a straight line AB, CD intersect at point E < DEA + < CEA = < CEA + < BEC = 180 ° (proposition 2) <

37、 DEA = < BEC Proposition 4: two straight line parallel, TongWeiJiao equal A linear EF and two parallel straight line AB, CD intersect Hypothesis is not equal to < GHD AGH < Might as well put < AGH larger < AGH + < BGH > < GHD + < BGH And < AGH + < BGH = 180 ° (p

38、roposition 1) < GHD + < BGH < 180 ° two straight line extension will intersect And two straight line parallel < AGH = < GHD And < AGH = < EGB (proposition 3) < GHD = < EGB Original proposition find Proposition 5: if two triangle, a two horns were equal to another two

39、horn, and side is equal to the other side, which have a side yes DengJiao home change, or is the DengJiao edge, then their other edge also equal to the other side, and the other to the horn of the horns of the other Proof: if AB indicates DE Might as well put AB > DE take BG is equal to DE Connec

40、tion GC BG = DE BC = EF GB = DE BC = EF < GBC = < DEF GC = DF And enables delta GBC enables delta DEF the rest Angle and edge also equal (proposition 1) < GCB = < DFE < BCG = < BCA It is not possible AB = DE And BC = EF AB = DE BC = EF < ABC = < DEF AC = DF < BAC = < ED

41、F (proposition 1) That indicates a EF BC Might as well put BC > EF Make BH = EF Link AH BH = EF AB = DE An Angle to equal AH = DF train ABH enables delta DEF < BHA = < EFD And < EFD = < BCA Therefore, in the triangle AHC, outside, BHA equal to < BCA It is not possible BC = EF And A

42、B = DE Angle are equal (proposition 1) enables delta ABC enables delta DEF AC = DF Proposition 6: in a parallelogram, edge is equal and diagonal halve its area (note: the geometric was the original text of the definition of no parallelogram Definition: in the same plane within two groups respectivel

43、y of the parallel quadrilateral called parallelogram. (1) if a quadrilateral is a parallelogram, so the two groups of side of quadrilateral are equal. (2) if a quadrilateral is a parallelogram, so the quadrilateral two sets of diagonal equal respectively. ) Proof: AB CD < ABC = < BCD AC BD <

44、; ACB = < CBD (proposition 4) BC = BC and enables delta ABC enables delta DCB < ABC = < BCD And < CBD = < ACB AC = AC enables delta ABD enables delta ACD < BAC = < CDB parallelogram ABCD, of the diagonal equal to each other (1), (2) properties have to card) Similarly, enables de

45、lta ABC enables delta DCB diagonal BC divide the area of the parallelogram ACBD Proposition 7: in the same base and in the same two parallel lines between the parallelogram equal Proof: set ABCD, EBCF is a parallelogram, they in the same bottom BC. And in the same parallel lines AF, between BC paral

46、lelogram ABCD is AD = BC Similarly, EF = BC, AD = EF AE = DF And AB = DC FDC = < EAB enables delta EAB enables delta FDC EB = FC area enables delta EAB-enables delta DGE = enables delta FDC-enables delta DGE area ABGD = EGCF With plus GBC accidents parallelogram ABCD area is equal to EBCF paralle

47、logram Proposition 8: if any straight line on a bit have two straight line is not this a straight line with side, and a straight line and LinJiao and equals two right angles, then these two straight lines in the same line Proof: if BD and BC of line BE and CB co-line hypothesis AB in straight lines

48、above CBE < ABC + < ABE = 180 ° (proposition 2) And ABC < + < ABD = 180 ° < CBA + < ABE = < CBA + < ABD Both sides also minus the CBA < The < ABE = < ABD (axiom 4, axiom 1, axiom 3) It is not possible BE, BC of line Similarly in addition to no other lines a

49、nd the BD BC were line CB and BD altogether line Proposition 9: in the same base and in the same between two parallel lines equal triangle area Proof: as shown in figure, ABC set triangle, with the same DBC and two parallel lines AD, between BC Extend the AD and DA respectively to F, E, and BE as parallel to the CA B, C for CF, parallel to the BD The EBCA and DBCF are quadrilateral parallelogram, and the area is equal (proposition 5) enables delta ABC is the area of the idol is must EBCA half Train DBC is the area of the parallelogram half the DBCF (proposition 6) enables delta area is