Fig. 22. Fig. 23. Fig. 24. fore falls upon the point F, and the two triangles coin. +...or " turned over, and flaced upon the triangle DEF, and because the bases are equal, and the angles at the bases the same, the side DF' would fall upon DF, and the side EF would fall upon EF : the point F" would therefore fall directly upon the point F : and the two triangles would entirely coincide in all their parts. We say, therefore—If two triangles have a side and the two adjacent angles of the one, equal to a side and the two adjacent angles of the other, each to each, the two triangles will be equal in all their parts; that is, the third angle of the one will be equal to the third angle of the other, and the two remaining sides of the one will be equal respectively to the two remaining sides of the other. This is the third case of equal triangles. 44. Problem. The hypothen use and one other side of a right-angled triangle being given, to construct the triangle. Y Let the line A (fig. 25) equal the hypothenuse, and the line B, the other given side of the triangle. Draw the line CD equal to the given side B ; and at D draw the indefinite line DE, perpendicular to CD ; then from C as a centre, with a radius equal to the given hypothen use, draw an arc cutting the perpendicular in E ; draw CE, and you have the right-angled triangle required. [Let the learner show that no different triangle could be constructed with these things given.] We say then—If two right-angled triangles have “he hypothen use and a side of the one, equal respectively to the hypothen use and a side of the other, the two triangles will be equal in all their parts. This is the fourth case of equal triangles. 45. We have seen that the sum of the three angles of every triangle, is equal to two right-angles; yet there is an infinite variety in the magnitude of the individual ansties; and this evidently depends upon the relative maginitudes of the sides. We have learned that when the sides are the same, the angles must be the same (39). Let us examine this subject a little farther. We will Fig. 25. take the isosceles triangle ABC (fig. 26), of which AB Fig. 26. and AC are the two equal sides. From the vertex A, to the middle of the base BC, draw the straight line AD. This will divide the triangle ABC, into two trian gles ADB, ADC. If we examine them, we shall see that AB of the one is equal to AC of the other, that DB 2 Fig. 26. Fig. 27. of the one is equal to DC of the other, and that AD is a e AB be the greater side ; take upon AB, the part AD equal to AC, and draw DC. The angle ADC may be considered as an exterior angle to the triangle DBC, and therefore greater than the angle B; the triangle ADC is isosceles, and consequently has the angle ACD equal to the angle ADC ; but the angle ACB is greater than ACD ; for a still stronger reason is the angle ACB, greater than ABC. But the side AB is greater than AD, or than its equal AC. We see then that the greater side is opposite to the greater angle. If we had compared either of these angles with the angle A, and also their opposite sides, the result would have been similar ; whence we have this general rule—In any triangle, the greater side is opposite to the greater angle. The side opposite to an angle is frequently said to subtend the angle. 48. If two sides of a triangle are unequal, the opposite angles will be unequal (47); therefore——If in any triangle two of the angles are equal, the sides opposite those angles will be equal, and the triangle will be isosceles. 49. In a right-angled triangle, as the two oblique angles must be each less than a right-angle, –the hypothenuse (the side opposite the right angle) must be greater than either of the other sides. The two sides which contain the right-angle are frequently called the legs of the triangle. 50. Either side of a triangle, being a straight line and therefore the shortest which can be drawn between the vertices at its extremities, must be less than the sum of the other two sides. It follows from this, that either side of a triangle must be greater than the difference between the other two sides; if this were not the case, the greatest would exceed or at least equal the sum of the two others. ... 51. Suppose AB (fig. 29) perpendicular to DC, and Fig. * AE and AF two lines drawn from the same point A, and falling obliquely upon DC at equal distances from the point B; the two triangles ABE and ABF, will have AB a common side, EB of the one equal to BF of the other, and the angle ABE equal to the angle ABF (11); the triangles will be equal by the second case, and will have the hypothen uses AE, AF equal, and each greater than the perpendicular AB (49). Draw AG at a greater distance from the perpendicular than AE is ; you will have the triangle AEG obtuse-angled at E, because this angle is the supplement of the acute angle AEB (28); and in the triangle AEG, the side AG being opposite to the obtuse angle is greater than AE opposite an acute angle ; and we have these three propositions. If from a point without a straight line, a perpendicular be drawn to that line, and also several oblique lines, – (1.) The perpendicular is less than either of the oblique lines. (2.) Oblique lines equally distant from the perpendicular are equal ; and (3.) Of two oblique lines, that is the greater which is at the greater distance from the perpendicular. 52. It follows from the second of these propositions, that—If a perpendicular be drawn through the middle of a line, A B, every point in the perpendicular will be equally distant from the two extremities of this line; for Fig. 29, Fig. 30. Fig. 31. straight lines drawn from any point in the perpendicular |