straight line, only, instead of being drawn in opposite directions from A as in the text, they will sometimes be drawn in the same direction; that s ABD and FBC will sometimes be supplementary instead of equal; and that then the equality of As ABD and FBC will follow, not from I. 4, but from I. 38, Cor. All the different varieties of figure are obtained thus : Call X the square on the hypotenuse, Y and Z the squares on the other sides. Describe (1) X outwardly, Y outwardly, Z outwardly. The following methods of exhibiting how two squares may be dissected and put together so as to form a third square, are probably the simplest and neatest ocular proofs yet given of this celebrated proposition: ABGH, BCEF are two squares placed side by side, and so that AB and BC form one straight line. Cut off CD = AB, and join ED, DH. (1) If, round E as a pivot, A ECD is rotated like the hands of a watch through a right angle, it will occupy the position EFK. If, round Has a pivot, △ HAD is rotated in a manner opposite to the hands of a watch through a right angle, it will occupy the position HGK. The two squares ABGH and BCEF will then be transformed into the square DEKH. (2) If ECD be slid along the plane in such a way that EC always remains vertical, and D moves along the line DH, it will come to occupy the position KGH. If △ HAD be slid along the plane in such a way that HA always remains vertical, and D moves along the line DE, it will come to occupy the position KFE. The two squares ABGH and BCEF will then be transformed into the square DEKΗ. [This method is substantially that given by Schooten in his Exercitationes Mathematicæ (1657), p. 111. The first or rotational way of getting as ECD, HAD into their places is given by J. C. Sturm in his Mathesis Enucleata (1689), p. 31; the second or translational way is mentioned by De Morgan in the Quarterly Journal of Mathematics, vol. i. p. 236.] ABC is a right-angled triangle. BCED is the square on the hypotenuse, ACKH and ABFG are the squares on the other sides. Find the centre of the square ABFG, which may be done by drawing the two diagonals (not shown in the figure), and through it draw two straight lines, one of which is || BC, and the other 1 BC. The square ABFG is then divided into four quadrilaterals equal in every respect. Through the middle points of the sides of the square BCED draw parallels to AB and AC as in the figure. Then the parts 1, 2, 3, 4, 5 will be found to coincide exactly with 1', 2', 3', 4', 5'. [This method is due to Henry Perigal, F.R.A.S., and was discovered about 1830. See The Messenger of Mathematics, new series, vol. ii. pp. 103–106.] 1. Show how to find a square = the sum of two given squares. 7. The square described on a diagonal of a given square is twice the given square. 8. Hence prove that the square on a straight line is four times the square on half the line. 9. The squares described on the two diagonals of a rectangle are together equal to the squares described on the four sides. 10. The squares described on the two diagonals of a rhombus are together equal to the squares described on the four sides. 11. If the hypotenuse and a side of one right-angled triangle be equal to the hypotenuse and a side of another right-angled triangle, the two triangles are equal in every respect. 12. If from the vertex of any triangle a perpendicular be drawn to the base, the difference of the squares on the two sides of the triangle is equal to the difference of the squares on the segments of the base. 13. The square on the side opposite an acute angle of a triangle is less than the squares on the other two sides. 14. The square on the side opposite an obtuse angle of a triangle is greater than the squares on the other two sides. 15. Five times the square on the hypotenuse of a right-angled triangle is equal to four times the sum of the squares on the medians drawn to the other two sides. 16. Three times the square on a side of an equilateral triangle is equal to four times the square on the perpendicular drawn from any vertex to the opposite side. 17. Divide a given straight line into two parts such that the sum of their squares may be equal to a given square. Is this always possible ? 18. Divide a given straight line into two parts such that the square on one of them may be double the square on the other. 19. If a straight line be divided into any two parts, the square on the whole line is greater than the sum of the squares on the two parts. 20. The sum of the squares of the distances of any point from two opposite corners of a rectangle is equal to the sum of the squares of its distances from the other two corners. The following deductions refer to the figure of the proposition in the text. They are all, or nearly all, given in an article in Leybourn's Mathematical Repository, new series, vol. iii. (1814), Part II. pp. 71-80, by John Bransby, Ipswich. 21. What is the use of proving that AG and AC are in the same straight line, and also AB and AH? 22. AF and AK are in the same straight line. 23. BG is || CH. 24. Prove As ABD, FBC equal by rotating the former round B through a right angle. Similarly, prove △s ACE, KCB equal. 25. Hence prove AD 1 FC, and AE 1 KB. 26. Ls ABC and DBF are supplementary, as also are 2s ACB and ECK. 27. Hence prove △s FBD, KCE = △ ABC. 28. FG, KH, LA all meet at one point T. 29. As AGH, THG, GAT, HTA are each = ∆ABC. 30. If from D and E, perpendiculars DU, EV be drawn to FB and KC produced, As UBD and VEC are each = △ ABC. 31. DF2 + EK2 = 5 BC2. 32. The squares on the sides of the polygon DFGHKE = 8 BC2. 33. If from Fand K perpendiculars FM, KN be drawn to BC produced, and I be the point where AL meets BC, A BFM = △ ABI, and △ CKN = ∆ ACI. 34. FM + KN = BC, and BN = CM = AL. 35. If DB and EC produced meet FG and KH at P and Q, prove by rotating A ABC that it = each of the As FBP, KCQ. 36. If PQ be joined, BCQP is a square. 37. ABPT is a ||m, and = rectangle BL; ACQT is a ||m, and = rectangle CL. 38. ADBT is a ||m, and = rectangle BL; AECT is a ||m, and = rectangle CL. 39. DFPU and EKQV are ||ms, and each = 4 △ ABC. 40. ADUH and AEVG are ||ms, and each = 2 △ ABC. 41. BK is CT, and CF 1 BT. 42. Hence prove that AL, BK, CF meet at one point O. (See App. I. 3.) 43. If BK meet AC in X, and CF meet AB in W, As BHX, CG W are each = △ABC. 44. AW AX. 45. ACW = △ BCX, and △ ABX = △ BCW. 46. Quadrilateral A WOX = ∆ BOC. 47. If from G and H perpendiculars GR, HS be drawn to BC or BC produced, and if these perpendiculars meet AB and AC in Y and Z, prove by rotating △ ABC that it = GAY or △ ZAH. 48. DU produced passes through Z, EV produced through Y, GV through W, and HU through X. 49. If through A a parallel to BC be drawn, meeting GR in G', and HS in H', As AGG', AZH' are = ∆ABI, and As AYG and AHH' = △ ACI. 50. IR = IS; GR + HS = MN; FM + GR + HS + KN = 2 (BC + AI); GR = BS; HS = CR. PROPOSITION 48. THEOREM. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by those two sides is a right angle. D Let ABC be a triangle, and let BC2 = BA2 + AC2 : it is required to prove L BAC right. From A draw AD 1 AC, and = AB; and join CD. Because AD = AB; .·. AD2 = AB2. To each of these equals add AC2; ... AD2 + AC2 = AB2 + AC2. I. 11, 3 |