1. In the construction it is said, draw AD 1 BC. Would it not be simpler, and answer the same purpose, to say, produce AB to D. Why? 2. Prove the proposition indirectly by drawing AD 1 AC, and on the same side of AC as AB, and using I. 7 (Proclus). 3. If the square on one side of a triangle be less than the sum of the squares on the other two sides, the angle opposite that side is acute. 4. If the square on one side of a triangle be greater than the sum of the squares on the other two sides, the angle opposite that side is obtuse. 5. Prove that the triangle whose sides are 3, 4, 5 is right-angled.* 6. Hence derive a method of drawing a perpendicular to a given straight line from a point in it. 7. Show that the following two rules,+ due respectively to Pythagoras and Plato, give numbers representing the sides of right-angled triangles, and show also that the two rules are fundamentally the same. (a) Take an odd number for the less side about the right angle. Subtract unity from the square of it, and halve the remainder; this will give the greater side about the right angle. Add unity to the greater side for the hypotenuse. (b) Take an even number for one of the sides about the right angle. From the square of half of this number subtract unity for the other side about the right angle, and to the square of half this number add unity for the hypotenuse. *This is said by Plutarch to have been known to the early Egyptians. + See Friedlein's Proclus, p. 428, and Hultsch's Heronis . . . reliquiæ, pp. 56, 57. APPENDIX I. PROPOSITION 1. The straight line joining the middle points of any two sides of a triangle is parallel to the third side and equal to the half of it. Let ABC be a triangle, and let L, K be the middle points of AB, AC: it is required to prove LK || BC and = Join BK, CL. half of BC. Because AL = BL, and because AK = CK, .. A BLC= = Δ ΒΚΟ. .. LK is || BC. I. 39 Hence, if H be the middle of BC, and HK be joined, HK is || AB; .. BHKL is a m; .. LK = BH = half of BC. I. Def. 33 COR. 1.-Conversely, The straight line drawn through the middle point of one side of a triangle parallel to a second side bisects the third side.* COR. 2.-AB is a given straight line, C and D are two points, either on the same side of AB or on opposite sides of AB, and such that AC and BD are parallel. If through E the middle point of AB, a straight line be drawn || AC or BD to meet CD at F, then * The corollaries and converses given in the Appendices should be proved to be true. Many of them are not obvious. F is the middle point of CD, and EF is equal either to half the sum of AC and BD, or to half their difference. PROPOSITION 2. The straight lines drawn perpendicular to the sides of a triangle from the middle points of the sides are concurrent (that is, pass through the same point). See the figure and demonstration of IV. 5. I. 8 If S be joined to H, the middle of BC, then SH is 1 BC. PROPOSITION 3. The straight lines drawn from the vertices of a triangle perpendicular to the opposite sides are concurrent.* Let AX, BY, CZ be the three perpendiculars from A, B, C on the opposite sides of the ▲ ABC; it is required to prove AX, BY, CZ concurrent. Through A, B, C draw KL, LH, HK || BC, CA, AB. Then the figures ABCK, ACBL are ||ms ; I. 31 I. Def. 33 I. 34 * Pappus, VII. 62. The proof here given seems to be due to F. J. Servois see his Solutions peu connues de différens problèmes de Géométriepratique (1804), p. 15. It is attributed to Gauss by Dr R. Baltzer. Hence also, B and C are the middle points of LH and HK. But since AX, BY, CZ are respectively 1 BC, CA, AB, they must be respectively 1 KL, LH, HK, and.. concurrent. Const. App. I. 2 NOTE.-The point O is called the orthocentre of the ▲ ABC (an expression due to W. H. Besant), and ▲ XYZ, formed by joining the feet of the perpendiculars, is called sometimes the pedal, sometimes the orthocentric, triangle. Let the medians BK, CL of the ▲ ABC meet at G: it is required to prove that, if H be the middle point of BC, the median AH will pass through G. that is, the median CL cuts BK at its point of trisection remote from B. Hence also, the median AH cuts BK at its point of trisection remote from B, that is, AH passes through G. COR.-If the points H, K, L be joined, the medians of the A HKL are concurrent at G. NOTE.-The point G is called the centroid of the ▲ ABC (an expression due to T. S. Davies), and ▲ HKL may be called the median triangle. The centroid of a triangle is the same point as that which in Statics is called the centre of gravity of the triangle, and may be found by drawing one median, and trisecting it. PROPOSITION 5. The orthocentre, the centroid, and the circumscribed centre of a triangle are collinear (that is, lie on the same straight line), and the distance between the first two is double of the distance between the last two.* Let ABC be a triangle, O its orthocentre determined by drawing AX and BY 1 BC and CA; S its circumscribed centre determined by drawing through H and K the middle points of BC and CA, HS and KSBC and CA; and AH the median from A: it is required to prove that if SO be joined, it will cut AH at the centroid. Let SO and AH intersect at G; join P and Q, the middle points of GA, GO; ОА, ов; Because H and K are the middle points of CB, CA; Because U and V are .. UV is || AB and .. HK is || UV and the middle points of OA, OB ; = half AB, = UV. App. I. 1 App. I. 1 * First given by Euler in 1765. See Novi Commentarii Academiæ Scientiarum Imperialis Petropolitanæ, vol. xi. pp. 13, 114. |