关于毕达哥拉斯定理证明的论文.doc

_关于毕达哥拉斯定理的证明专业：姓名：指导老师：摘要：对于几何原本中毕达哥拉斯定理的证明过程，欧几里得以定义，公设，公理的方式进行推理，现将所有涉及毕达哥拉斯定理的证明命题提出。关键词：毕达哥拉斯定理，定义，公设，公理。正文：定义：1. 点是没有大小的东西2.线只有长度而没有宽带3.一线的两端是点4.直线是它上面的点一样地平放着的线5.面只有长度和宽带6.面的边缘是线7.平面是它上面的线一样地平放着 8. 平面角是在一平面内但不在一条直线上的两条相交线相互的倾斜度.9. 当包含角的两条线都是直线时，这个角叫做直线角.10. 当一条直线和另一条直线交成邻角彼此相等时，这些角每一个被叫做直角，而且称这一条直线垂直于另一条直线。11. 大于直角的角称为钝角。12. 小于直角的角称为锐角13. 边界是物体的边缘14. 图形是一个边界或者几个边界所围成的15. 圆：由一条线包围着的平面图形，其内有一点与这条线上任何一个点所连成的线段都相等。16. 这个点（指定义15中提到的那个点）叫做圆心。17. 圆的直径是任意一条经过圆心的直线在两个方向被圆截得的线段，且把圆二等分。18.半圆是直径与被它切割的圆弧所围成的图形，半圆的圆心与原圆心相同。19.直线形是由直线围成的.三边形是由三条直线围成的,四边形是由四条直线围成的,多边形是由四条以上直线围成的.20.在三边形中,三条边相等的,叫做等边三角形;只有两条边相等的,叫做等腰三角形;各边不等的,叫做不等边三角形.21.此外,在三边形中,有一个角是直角的,叫做直角三角形;有一个角是钝角的,叫做钝角三角形;各边不等的,叫做不等边三角形.22.在四边形中,四边相等且四个角是直角的,叫做正方形;角是直角,但四边不全相等的,叫做长方形;四边相等,但角不是直角的,叫做菱形;对角相等且对边相等,但边不全相等且角不是直角的,叫做斜方形;其余的四边形叫做不规则四边形.23.平行直线是在同一个平面内向两端无限延长不能相交的直线.0公理：1.等于同量的彼此相等2.等量加等量，其和相等；3.等量减等量，其差相等4.彼此能重合的物体是全等的5.整体大于部分。公设:1.过两点能作且只能作一直线；2.线段(有限直线)可以无限地延长；3.以任一点为圆心,任意长为半径,可作一圆；4.凡是直角都相等； 5.同平面内一条直线和另外两条直线相交，若在直线同侧的两个内角之和小于180，则这两条直线经无限延长后在这一侧一定相交。作图证明：1.在一个已知有限直线上作一个等边三角形设AB是已知直线以A为圆心，以AB为距离画圆以B为圆心，以AB为距离画圆两圆交点C到A,B的来连线CA,CBAC=ABBC=BACA=CB=ABABC是等边三角形2.过直线外一已知点作一直线平行于已知直线。设A是已知点，BC是已知直线，要求经过A点做直线平行于BC在BC上任取一点D，连接AD，在直线DA上的点A，做 DAE=ADC设直线AF是直线EA的延长线直线AD和两条直线BC,EF相交成彼此相等的内错角EAD，ADC EAFBC作毕3.在已知线段上作一个正方形。设AB是已知线段，要求在线段AB上作一个正方形令AC是从线段AB上的点A所画的直线，它与AB成直角取AD=AB过点D做DE平行于AB，过点B做BE 平行于AD，所以ADEB是平行四边形AB=DE,AD=BE又AD=AB平行四边形ADEB是等边的BAD+ADE=180BAD 是直角ADE是直角平行四边形中对边及对角相等ABDE是正方形 4：由已知直线上一已知点做直线与已知直线成直角解：设在AC上任意取一点D，使CE=CD在DE上作一个等边三角形FDE连接FCDC=CECF=CFDF=CFDF=FEDCF=ECF他们是邻角，由定义10，二者都是直角作毕。5：已知两条不相等的线段，试由大的上边截取一条线段是它等于另外一条设AB，C是两条不相等的线段由A取AD等于线段C以A为圆心，AD为距离画圆DEFA是圆DEF的圆心AE=AD又C=ADAE=C=AD作毕命题证明：命题1：如果两个三角形有两边分别等于两边，而且这些相等的线段所夹的角相等。那么，它们的底边等于底边，三角形全等于三角形，而且其它的角等于其它的角，即那等边所对的角。证明：设ABC,DEF是两个三角形，AB=DE,AC=DF， BAC=EDF如果移动三角形ABC到DEF上，若A落在点D上，且线段落在DE上AB=DEB与E重合又AB与DE重合 BAC=EDFAC与DF重合又AC=DFC与F重合ABC与DEF重合，即全等命题2：一条直线和另一条直线所交成的角，或者是两个直角，或者是它们的和等于2个直角证明：设任意直线AB交CD成角CBA,ABD若CBA=ABD则CBA=ABD=90（定义10）若二者不是直角作BECD于BCBE=EBD=90CBE=CBA+ABECBE+EBD=CBA+ABE+EBD同理，DBA+ABC=DBE+EBA+ABCCBE+EBD=DBA+ABC=180原命题得证命题3：对顶角相等证明：设直线AB,CD相交于点EDEA+CEA=CEA+BEC=180（命题2）DEA=BEC命题4：两直线平行，同位角相等设直线EF与两条平行直线AB,CD相交假设AGH不等于GHD不妨设AGH较大AGH+BGHGHD+BGH又AGH+BGH=180（命题1）GHD+BGH180二直线延长一定会相交又两直线平行AGH=GHD又AGH=EGB（命题3）GHD=EGB原命题得证命题5：如果在两个三角形中，一个的两个角分别等于另一个的两个角，而且一边等于另一个的一边，即过着这边是的等角的家变，或者是等角的对边，则它们的其他的边也等于其他的边，且其他的角也等于其他的角证明：如果ABDE不妨设ABDE取BG等于DE连接GCBG=DEBC=EFGB=DEBC=EFGBC=DEFGC=DF又GBCDEF其余角和边也相等（命题1）GCB=DFEBCG=BCA这是不可能的AB=DE又BC=EFAB=DEBC=EFABC=DEF AC=DFBAC=EDF（命题1）假设BCEF不妨设BCEF令BH=EF连接AHBH=EFAB=DE所成的夹角相等AH=DFABHDEFBHA=EFD又EFD=BCA因此，在三角形AHC中，外角BHA等于BCA这是不可能的BC=EF又AB=DE夹角也相等（命题1）ABCDEFAC=DF命题6：在平行四边形中，对边相等且对角线二等分其面积（注：几何原本原文中无平行四边形的定义定义: 在同一平面内两组对边分别平行的四边形叫做平行四边形。（1）如果一个四边形是平行四边形，那么这个四边形的两组对边分别相等。（2）如果一个四边形是平行四边形，那么这个四边形的两组对角分别相等。）证明：ABCDABC=BCDACBDACB=CBD（命题4）又BC=BCABCDCBABC=BCD又CBD=ACBAC=ACABDACDBAC=CDB平行四边形ABCD中，对边对角彼此相等（1）（2）性质得证）同样地，ABCDCB对角线BC平分平行四边形ACBD的面积命题7：在同底且在相同两平行线之间的平行四边形面积相等证明：设ABCD，EBCF是平行四边形，它们在同底BC。且在相同的平行线AF，BC之间ABCD是平行四边形AD=BC同理，EF=BC,AD=EFAE=DF又AB=DCFDC=EABEABFDCEB=FC面积EAB-DGE=FDC-DGE面积ABGD=EGCF同加上GBC平行四边形ABCD面积等于平行四边形EBCF命题8：如果过任意一条直线上一点有两条直线不在这一直线的同侧，且和直线所成邻角和等于二直角，则这两条直线在同一条直线上证明：如果BD与BC不共线假设BE和CB共线AB在直线CBE之上ABC+ABE=180（命题2）又ABC+ABD=180CBA+ABE=CBA+ABD两边同时减去CBA则ABE=ABD（公设4，公理1，公理3）这是不可能的BE，BC不共线同理除BD外没有其他直线与BC共线CB与BD共线命题9：在同底上且在相同两平行线之间的三角形面积相等证明：如图所示，设三角形ABC,DBC同底且在相同两平行线AD，BC之间延长AD和DA分别至F，E，过B作BE平行于CA，过C作CF平行于BD则四边形EBCA和DBCF都是平行四边形，且面积相等（命题5）ABC的面积是偶像是必须EBCA的一半 DBC的面积是平行四边形DBCF的一半（命题6）DBC面积等于ABC的面积命题10：如果一个平行四边形和一个三角形既通敌又在两平行线之间，则平行四边形的面积是三角形的2倍证明：连接ACABC与EBC又同底BC，又在平行线BC和AE之间ABC的面积等于EBCAC平分平行四边形ABCD平行四边形ABCD的面积是EBC的2倍平行四边形ABCD的面积是EBC的2倍关于毕达哥拉斯定理的证明：直角三角形的直角边的平方和等于斜边的平方。已知：如图所示，ABC是直角三角形。求证：AB+AC=BC。证明：分别以直角边AB,AC和斜边BC的作正方形ABFG，正方形ACKH,正方形BCED；（作图3） 过A作AL平行于BD或CE，连接AD，FC； BAC=BAG=90 C,A,G共线（命题8） 同理，B,A,H共线 DBC=FBA 所以DBC+ABC=FBA+ABC 即DBA=FBC（公理2） 又DB=BC FB=BA 所以ABDFBC（命题1） 平行线AL与BD之间 平行四边形BL的面积是ABD的2倍 同理，正方形GB的面积是FBC的2倍 由公理2，平行四边形BL的面积与正方形BD相等（命题10） 同理可得，平行四边形CL等于正方形HC 正方形BCED的面积等于正方形ABFG与正方形ACKH面积之和（公理2） BC=AB+AC 原命题得证参考文献：欧几里得几何原本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 theorem 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 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 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. 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 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 and 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 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 equilateral 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 and 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 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 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 after 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 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, connection 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 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 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 FC 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, 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 And 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 angles 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 Similarly, 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) 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 + GHD + BGH And AGH + BGH = 180 (proposition 1) GHD + BGH 180 two straight line extension will intersect And two straight line parallel AGH = GHD And AGH = EGB (proposition 3) GHD = DE take BG is equal to DE Connection 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 = 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 AB = 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 respectively 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 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 delta 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 parallelogram 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 parallelogram 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 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 and 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 equal to train the DBC ABC area Proposition 10: if a parallelogram