? Place A ABC on A DEF, so that A is on D, and AB along DE, then B falls on E, :: AB = DE; and BC must lie along EF, : _B = LE, and :: pt. C must fall somewhere on EF. If pt. C does not fall on F, let it fall as at G; then _DGE is 2 ACB, and LACB = LDFE (given). :: _DGE = LDFE, which is impossible (I. 16). :: C must fall on F, :: ABC coincides with A DEF. :: BC = EF, AC = DF, and LA = _ D. Q.E.D. On Triangles equal to one another in every Respect. Triangles have six parts, three sides and three angles; and when the six parts of one triangle are respectively equal to the six parts of another, the triangles are said to be congruent, or equal in every respect. In general, when three independent parts of one triangle are known to be equal to the corresponding parts of another triangle, the two triangles can be proved equal to one another in every respect. The different cases of three parts respectively equal are these : 1. The three sides, I. 8. 2. Two sides and the contained angle, I. 4. 3. Two sides and an angle adjacent to one of them (called “the ambiguous case”), given in one form below as Ex. 64. 4. One side and two angles, both adjacent to the given side, I. 26, 1st Case. 5. One side and two angles, one only adjacent to the given side, I. 26, 2nd Case. 6. The three angles. These are not independent data (see I. 32), and triangles having their angles respectively equal are not necessarily equal in other respects. Since three given parts of a triangle, as in 1, 2, 4, and 5, can belong to only one triangle, or to any number of triangles identically equal to one another, a triangle is said to be determined when such three parts are given. 64. Two triangles are equal in every respect which have one angle equal in each, the two sides about another angle equal, and the third angles of the same kind, acute or obtuse, in each triangle (can be proved by I. 26). On Parallel Lines. The angles made by one straight line intersecting two others have received particular names from their relative positions. The angles 3 and 5 are called alternate angles ; 4 and 6 are also alternate. 2 and 6 are corresponding angles, or 2 2 exterior, and - 6 interior and opposite on same side of the intersecting line. 1 and 5, 8 and 4, 7 and 3, are named as 2 and 6. 3 and 6, also 4 and 5, are two interior angles on the same side of the intersecting line. 43 B F PROP. 27. THEOR. If a straight line falling upon two other straight lines make the alternate angles equal to one another, these two straight lines shall be parallel. Given EF meeting AB and CD, A E making AEF = _EFD. To prove AB || CD. C If AB be not || CD, they will meet if produced; let them be produced and meet at G. Then GEF will be a A, and AEF, the exterior angle, will be > EFG (1. 16). But L AEF is also = _ EFG(hyp.), which is impossible. :: AB and CD do not meet when produced towards B, D. In the same way it can be shown that they do not meet when produced towards A, C. :, AB and CD are parallel (Def. 28). Q.E.D. PROP. 28. THEOR. If a straight line falling on two other straight lines make the corresponding angles equal, the lines shall be parallel ; or, If a straight line falling on two other straight lines make the two interior angles on the same side of it equal to two right angles, the lines shall be parallel 1. Given EH meeting AB and CD, making _ EFB A B = 2 FGD, or any other angles equal. _ FGD (hyp.), = :: L AFG = LFGD, and these are alternate angles; :: AB is || CD (I. 27). 2. Given EH meeting AB and CD, making 2 BFG + LFGD 2 rt. angles. To prove AB || CD. :: _ AFG +_ BFG 2 L's. (I. 13), and 2 BFG + LFGD 2 L's. (hyp.), .. AFG + 2 BFG : = LBFG + _FGD(Ax. 1&10). Take away the common angle BFG, and there remains - AFG = 2 FGD, and these are alternate angles; :: AB || CD (1. 27). Q.E.D. 65. Perpendiculars to the same line are parallel to one another. 66. Every rectangular four-sided figure has its opposite sides parallel. 67. A rhombus and a rhomboid have their opposite sides parallel. PROP. 29. THEOR. F B G If a straight line fall upon two parallel straight lines, it makes the alternate angles equal, the corresponding angles equal, and the two interior angles on the same side of it together equal to two right angles. Given EH meeting the parallel lines AB and K 1. / AFG = _FGD. 3. _ BFG + LFGD = 2 LI's. 1. If AFG is not - FGD, make at F in FG Ľ KFG _FGD (I. 23). : _KFG LFGD, and these are alternate angles, :. KL is || CD (1. 27). :. AB and KL, two intersecting lines, are both parallel to a third, which is impossible (Ax. 11). :: L AFG is not unequal to · FGD, :: AFG = _FGD. 2. L AFG= _FGD (1st part), and – AFG = = L EFB (1. 15), :: LEFB = _ FGD (Ax. 1). 3. . AFG = LFGD (1st part), add - BFG to each. : . AFG + LBFG = L BFG + LFGD. But AFG + L BFG 2 Ls. (I. 13); ::BFG + FGD 2 L's. (Ax. 1). Q.E.D. 68. If two intersecting straight lines are parallel respectively to two other straight lines, the angle contained by the first two shall be equal to the angle contained by the other two. = CoR.-If a straight line meeting two other straight lines make the two interior angles on the same side less than two right angles, the two straight lines, if produced, will meet on that side on which are the angles less than two right angles. If _KFG +. FGC are í к To prove KF and CG will meet if produced towards K, C. If not, then KL is | CD, or they meet on the other side. KL is not || CD, for then _ KFG + LFGC would be 2 L'.s., which they are not. Neither do KL and CD meet towards L, D; for then - LFG and 2 FGD, being two angles of a A, would be < 2 Ls., and so, by I. 13, _ KFG + FGC would be > 2 L'.s., which they are not. :: KF and CG meet if produced towards K, C. Q.E.D. 69. If a straight line intercepted between two parallel lines is bisected, any other straight line drawn through the point of bisection to meet the parallel lines is also bisected at that point; and the two intersecting lines cut off equal portions of the parallel lines. 70. Given two parallel straight lines and a line perpendicular to one of them, it shall be perpendicular to the other. 71. In the isosceles triangle ABC draw AD perpendicular to the base BC, and through any point P in AD draw QPR at right angles to AD, meeting AB and AC in Q and R. Prove that AQR is an isosceles triangle. 72. AB is the hypotenuse of a right-angled triangle ABC, BD is drawn bisecting angle B meeting AC at D, and DE is drawn at right angles to AC meeting AB at E. Prove that EDB is an isosceles triangle. |