the angle ACB [I. 5]; but it is not [Hyp.]; therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would be less than the angle ACB [I. 18]; but it is not [Hyp.]; therefore AC is not less than AB. And it has been proved that AC is not equal to AB. Therefore AC is greater than AB. Therefore, the greater angle, &c. Ex. 1. Q.E.D. The perpendicular is shorter than any other line which can be drawn from a given point to a given line. 2. Any point is taken in the side of a square. Show that the sum of its distances from the four angles is less than the perimeter of the square. PROPOSITION 20. THEOREM. Any two sides of a triangle are together greater than the third side. 2. Produce BA to the point D, making AD equal to AC [I. 3], and join DC. Then, because AD is equal to AC [Const.], the 3. angle ADC is equal to the angle ACD. [1.5.] But the angle BCD is greater than the angle ACD. [Ax. 9.] Therefore the angle BCD is greater than the angle BDC. And because the angle BCD of the triangle BCD is greater than its angle BDC, and that the greater angle is subtended by the greater side [I. 19]; therefore the side BD is greater than the side BC. But BD is equal to BA and AC. Therefore BA, AC are greater than BC. In the same manner it may be proved that AB, BC are greater than AC, and BC, CA greater than AB. Therefore, any two sides, &c. Q.E.D. Ex. 1. Prove that the sides BC, CA are greater than AB. 2. If BA is greater than AC, cut off AE=AC, and show that BE the difference of AB and AC—is less than the third side BC. 3. The four sides of a quadrilateral are together greater than its two diagonals. PROPOSITION 21. THEOREM. If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let ABC be a triangle, and from the points 1. B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to the point D within the triangle: BD, DC shall be less than the other two sides BA, AC of the triangle, but shall contain an angle BDC greater than the angle BAC. 2. Produce BD to E. Because two sides of a triangle are greater than 3. the third side, the two sides BA, AE of the triangle ABE are greater than the side BE. [I. 20.] To each of these add EC. Therefore BA, AC are greater than BE, EC. Again; the two sides CE, ED of the triangle CED are greater than the third side CD. [I. 20.] To each of these add DB. Therefore CE, ÈB are greater than CD, DB. But it has been proved that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC. Again, because the exterior angle of any triangle 3. is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than the angle CED. [I. 16.] For the same reason, the exterior angle CEB of the triangle ABE is greater than the angle BAE; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends, &c. Q.E.D. Unless the lines are drawn from "the ends of the side of a triangle," the proposition will not necessarily be true. Similarly it may be shown that if two quadrilaterals stand on the same base, and one is enclosed by the other, the perimeter of the inner figure is less than that of the outer figure. Ex. If O be a point within the triangle ABC, prove that the sum of AO, BO, and CO is less than the sum of the three sides of the triangle. PROPOSITION 22. PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. Let A, B, C be the three given straight lines, of 1. which any two whatever are greater than the third; namely, A and B greater than C; A and C greater than B; and B and C greater than A: it is required to make a triangle of which the sides shall be equal to A, B, C, each to each. A. B C 2. Take a straight line DE terminated at the point D, but unlimited towards E, and make DF equal to A, FG equal to B, and GH equal to C. [I. 3.] From the centre F, at the distance FD, describe the circle DKL. [Post. 3.] From the centre G, at the distance GH, describe the circle HKL. Join KF, KG. The triangle KFG shall have its sides equal to the three straight lines A, B, C. 3. Because the point F is the centre of the circle DKL, FD is equal to FK. [Def. 15.] But FD is equal to A. [Const.] Therefore FK is equal to A. [Ax. 1.] Again, because the point G is the centre of the circle HLK, GH is equal to GK. [Def. 15.] But GH is equal to C. [Const.] Therefore GK is equal to C. [Ax. 1.] And FG is equal to B. [Const.] Therefore the three straight lines KF, FG, GK are equal to the three A, B, C; and therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. Therefore a triangle has been described, &c. Q.E.F. If any two of the three lines A, B, and C were not greater than the third, the construction would fail. Thus if B and C were not greater than A, then FH would not be greater than FD, and the circle described from centre F, with radius FD, would therefore pass through the point H, or fall beyond it. In either case one circle would lie wholly within the other, and there would be no point of intersection. In like manner the construction would fail if A and C were not greater than B, for then each circle would lie wholly without the other. Ex. 1. If the lines A, B, and C are all equal to one another, show that this proposition is identical with prop. 1. 2. Having given a side and one of the diagonals of a rhombus, show how to construct it. PROPOSITION 23. PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the 1. given point in it, and DCE the rectilineal angle; it is required to make at the given point A, in the given straight line AB, an angle equal to the given rectilineal angle DCE. 2. A In CD, CE take any points D, E, and join DE. Make the triangle AFG the sides of which shall be equal to the three straight lines CD, DE, EC; so that AF shall be equal to CD, AG to CE, and FG to DE. [I. 22.] The angle FAG shall be equal to the angle E F B and the base FG equal to the base DE [Const.]; therefore the angle FAG is equal to the angle DCE. [I. 8.] Therefore at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE. Q.E.F. Ex. 1. Show how to make an angle double of a given angle. 2. Construct a triangle, having given two of its sides and the angle included by them. |