关于毕达哥拉斯定理证明的论文

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,CB∵AC=ABBC=BA∴CA=CB=AB∴△ABC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∴EAF∥BC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=180°∠BAD 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FC∵DC=CECF=CFDF=CFDF=FE∴∠DCF=∠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DEF∵A是圆DEF的圆心∴AE=AD又C=AD∴AE=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=DE∴B与E重合又AB与DE重合∠BAC=∠EDF∴AC与DF重合又AC=DF∴C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BE⊥CD于B∠CBE=∠EBD=90°∠CBE=∠CBA+∠ABE∴∠CBE+∠EBD=∠CBA+∠ABE+∠EBD同理，∠DBA+∠ABC=∠DBE+∠EBA+∠ABC∴∠CBE+∠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E∵∠DEA+∠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+∠BGH>∠GHD+∠BGH又∠AGH+∠BGH=180°（命题1）∴∠GHD+∠BGH<180°∴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AB≠DE不妨设AB＞DE取BG等于DE连接GC∵BG=DEBC=EFGB=DEBC=EF∴∠GBC=∠DEFGC=DF又∵△GBC≌△DEF∴其余角和边也相等（命题1）∴∠GCB=∠DFE∴∠BCG=∠BCA这是不可能的∴AB=DE又BC=EF∴AB=DEBC=EF∠ABC=∠DEF∴AC=DF∠BAC=∠EDF（命题1）假设BC≠EF不妨设BC＞EF令BH=EF连接AH∵BH=EFAB=DE所成的夹角相等∴AH=DF∴△ABH≌△DEF∴∠BHA=∠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）∴△ABC≌△DEF∴AC=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：∵AB∥CD∴∠ABC=∠BCD∵AC∥BD∴∠ACB=∠CBD（命题4）又BC=BC∴△ABC≌△DCB∴∠ABC=∠BCD又∵∠CBD=∠ACBAC=AC∴△ABD≌△ACD∴∠BAC=∠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，∵△ABC≌△DCB∴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=EF∴AE=DF又AB=DCFDC=∠EAB∴△EAB≌△FDCEB=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=180°∴∠CBA+∠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AC∵△ABC与△EBC又同底BC，又在平行线BC和AE之间∴△ABC的面积等于△EBC∵AC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：AB²+AC²=BC²。

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=BCFB=BA所以△ABD≌△FBC（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）∴BC²=AB²+AC²原命题得证参考文献：欧几里得《几何原本》The proof of the Pythagorean theorem aboutProfessional: ××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 things2. Line length and not only broadband3. A at both ends of the line is the point4. Straight line is on it to the point of being the same line5. Faces only length and broadband6. The edge is line7. 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 Angle13. The boundary is the edge of the object14. The figure is a boundary or surrounded by several boundary15. 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. 0Justice: 1. Equal to about the same amount of equal to each other2. Add amount equal, its and equal;3. Reduced amount equal, the poor are equal4. Each other can overlap object is congruent5. 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 triangleSet AB is known straight lineWith A to the right, to draw circles AB distanceWith B to the right, to draw circles AB distanceTwo round) to A C, B to attachment of CA, CB∵AC = ABBC = BA∴CA = CB = AB∴enables delta ABC is an equilateral triangle2. 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 = < ADCA straight line is straight line EA AF∴linear AD and two straight lines BC, EF into each other NaCuoJiao intersection equal EAD, ADC∴EAF ∥BC3.In line for a known on the square.Line AB is a known, in the line AB requirements on a squareThe line AB to AC from point A are painting of the straight line, it and AB, at right anglesTake AD = ABLead 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 = BEAnd 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 square4: 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 = CDIn DE make one FDE equilateral triangleConnection FC∵DC CECF = CFDF = CFDF = FE∴< DCF = < ECFThey are LinJiao, by definition 10, both is right anglesProposition 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 = < EDFIf mobile triangle ABC to DEF, if A fall in point D, and line in the paragraph DE∵AB = DE∴ B and E coincidenceAnd AB and DE superposition< BAC = < EDF∴AC and DF superpositionAnd AC = DF∴ C and F coincidence∴enables delta ABC and train DEF coincidence, that is congruentProposition 2: a straight line and the other a straight line pay into horn, or two right angles, or istheir and equal to two right anglesProof: set any straight line AB/CD into Angle CBA, ABDIf < CBA = < ABDThe < CBA = < ABD = 90 ° (definition 10)If both not right anglesBE as an CD in B< CBE = < EBD = 90 °< CBE = < CBA + < ABE∴< CBE + EBD < = < CBA + < ABE + < EBDSimilarly, < DBA + < ABC = < DBE + < EBA + < ABC∴< CBE + EBD < = < DBA + < ABC = 180 °Original proposition findProposition 3: vertical angles equalProof: a straight line AB, CD intersect at point E∵< DEA + < CEA = < CEA + < BEC = 180 ° (proposition 2) ∴< DEA = < BECProposition 4: two straight line parallel, TongWeiJiao equalA 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 + < BGHAnd < AGH + < BGH = 180 ° (proposition 1)∴< GHD + < BGH < 180 °∴two straight line extension will intersectAnd two straight line parallel∴< AGH = < GHDAnd < AGH = < EGB (proposition 3)∴< GHD = < EGBOriginal proposition findProposition 5: if two triangle, a two horns were equal to another two 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 DEMight as well put AB > DE take BG is equal to DEConnection GC∵BG = DEBC = EFGB = DEBC = EF∴< GBC = < DEFGC = DFAnd ∵enables delta GBC ≌enables delta DEF∴the rest Angle and edge also equal (proposition 1)∴< GCB = < DFE∴< BCG = < BCAIt is not possible∴AB = DEAnd BC = EF∴AB = DEBC = EF< ABC = < DEF∴AC = DF< BAC = < EDF (proposition 1)That indicates a EF BCMight as well put BC > EFMake BH = EFLink AH∵BH = EFAB = DEAn Angle to equal∴AH = DF∴train ABH ≌enables delta DEF∴< BHA = < EFDAnd < EFD = < BCATherefore, in the triangle AHC, outside, BHA equal to < BCA It is not possible∴BC = EFAnd AB = DEAngle are equal (proposition 1)∴enables delta ABC ≌enables delta DEF∴AC = DFProposition 6: in a parallelogram, edge is equal and diagonal halve its area (note: the geometric was the original text of the definition of no parallelogramDefinition: 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 = < BCDAnd ∵< CBD = < ACBAC = 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 ACBDProposition 7: in the same base and in the same two parallel lines between the parallelogram equalProof: 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 = BCSimilarly, EF = BC, AD = EF∴AE = DFAnd AB = DCFDC = < EAB∴enables delta EAB ≌enables delta FDCEB = FC∴area enables delta EAB-enables delta DGE = enables delta FDC-enables delta DGE∴area ABGD = EGCFWith plus GBC accidents∴parallelogram ABCD area is equal to EBCF parallelogramProposition 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 lineProof: if BD and BC of lineBE and CB co-line hypothesis∵AB in straight lines above CBE∴< ABC + < ABE = 180 ° (proposition 2)And ABC < + < ABD = 180 °∴< CBA + < ABE = < CBA + < ABDBoth sides also minus the CBA <The < ABE = < ABD (axiom 4, axiom 1, axiom 3)It is not possible∴BE, BC of lineSimilarly in addition to no other lines and the BD BC were line∴CB and BD altogether lineProposition 9: in the same base and in the same between two parallel lines equal triangle areaProof: as shown in figure, ABC set triangle, with the same DBC and two parallel lines AD, between BCExtend the AD and DA respectively to F, E, and BE as parallel to the CA B, C for CF, parallel to the BDThe 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 halfTrain DBC is the area of the parallelogram half the DBCF (proposition 6)∴enables delta area is equal to train the DBC ABC areaProposition 10: if a parallelogram and a triangle is collaborating again in two parallel lines between, is the area of a parallelogram is a triangle 2 timesProof: connect AC∵enables delta ABC and train EBC and with bottom BC, and in parallel lines BC and AE between ∴train the area of the ABC is equal to train EBC∵AC divide the parallelogram ABCD∴parallelogram ABCD is the area of the train EBC twice∴parallelogram ABCD is the area of the train EBC twiceThe proof of the Pythagorean theorem about:Right side of a right triangle hypotenuse is equal to the sum of the square.The known: as shown in figure, train ABC is a right triangle.Confirmed: AB ² + AC ² = BC ².Proof: respectively by orthogonal edge AB, AC and tapered side plain wheels of BC as a square ABFG, square ACKH, square BCED; (graphic 3)Over A parallel to the BD or for AL CE, connection AD, FC;∵< BAC = < BAG = 90 °∴C, A, G (proposition 8) were lineSimilarly, B, A, H of line∵< DBC = < FBASo < DBC + < ABC = < FBA + < ABCNamely < DBA = < FBC (justice 2)And DB = BCFB = BASo enables delta ABD ≌enables delta FBC (proposition 1)Parallel lines between AL and BDThe area of the parallelogram BL is 2 times of ABD accidentsSimilarly, a square GB is the area of the train FBC twice2 by justice, the area of the parallelogram BL and square equal BD (proposition 10)Similarly, a parallelogram CL is equal to the square HC∴square BCED area is equal to the square ABFG and square the size of ACKH (justice 2)∴BC ² = AB ² + AC ²Original proposition findReference: kansai the geometric originally。