42. Given the difference between the segments of the base made by the perpendicular, the sum of the squares of the sides, and the area; to construct the triangle. 43. Given the base, one of the angles at the base, and the differ- ence between the side opposite to it and the perpendiculas; to 44. Given the vertical angle, the difference of the base and one side, and the sum of the perpendicular drawn from the angle at the base contiguous to that side upon the opposite side and the segment cut off by it from that opposite side contiguous to the other angle at the base ; to construct the triangle. 45. Given the base, the difference of the sides, and the segment intercepted between the vertex and a perpendicular from one of the angles at the base upon the opposite side; to construct the triangle. 46. Given the vertical angle, the side of the inscribed square, and the rectangle contained by one side and its segment adjacent to the. base made by the angular point of the inscribed square; to GEOMETRICAL PROBLEMS. Angles and straight lines SECT. I. B (1.) From a given point, to draw the shortest line possible to a given straight line. Let A be the given point, and BD the given line. From A let fall the perpendicular AC; this will be less than any other line AD drawn from A to BD. For since AC is perpendicular to BD, the angle ACD is a right angle, therefore the angle ADC is less than a right angle (Eucl. i. 32.) and consequently less than ACD. But the greater angle is subtended by the greater side (Eucl. i. 19.); therefore AD is greater than AC. In the same manner every other line drawn from A to BD may be shewn to be greater than AÇ; therefore AC is the least. (2.) If a perpendicular be drawn bisecting a given straight line; any point in this perpendicular is at equal A E D distances, and any point without the perpendicular is at unequal distances from the extremities of the line. From C the point of bisection let Join AD, DB. Since AC = CB and CD is common, and the angle ACD=BCD being right angles, AD = DB. And the same may be proved of lines drawn from any other point in CD to A and B. But if a point E be taken which is not in CD, join EA cutting the perpendicular in D; join EB, DB. Then AD=DB from the first part, and AE is equal to AD, DE, that is, to BD, DE, and is therefore greater than BE, (Eucl. i. 20.); therefore, &c. A B B D P (3.) Through a given point to draw a straight line which shall make equal angles with two straight lines given in position. Let P be the given point, and BE, CF the lines given in position. Produce BE, CF to meet in A, and bisect the angle BAC by the line AD. From P let fall the perpendicular PD, and produce it both ways to E and F. It will be the line required. For the angle EAD is equal to the angle FAD, the angles at D right angles, and AD common, therefore (Eucl. i. 26.) the angle AED is equal to the angle AFD; therefore, &c. с B (4.) From two given points to draw two equal straight lines which shall meet in the same point of a line given in position. Let A and B be the given points, and CD the given straight line. Join AB, and bisect it in F, and from F draw FE at right angles to AB meeting CD in E, E is the point required. Join AE, EB. Since AF= FB, and FE is common, and the angles at F are right angles, therefore AE = EB. F 'E D B (5.) From two given points on the same side of a line given in position, to draw two lines which shall meet in that line, and make equal angles with it. "В Let A and B be the given points Since AD=DC and DP is common, and the angles at D are right angles, therefore the triangles APD, CPD are equal, and the angle APD = CPD = the vertically opposite angle BPE. (6.) From two given points on the same side of a line given in position, to draw two lines which shall meet in a point in this line, so that their sum shall be less P E than the sum of any two lines drawn from the same points and terminated at any other point in the same line. Let A and B be the given points, and DE the line given in position; from A and B let fall the perpendiculars AD, BE, and produce AD to C making CD = DA. Join BC cutting DE in P. Join AP; AP and PB shall be less than any other two lines Ap, pB drawn from A and B to any other point p in the line DE. For AD=DC and DP is common and the angles at D are right angles, .. AP= PC. In the same manner, if pС be joined, it may be shewn that Ap=pC. Hence AP and BP together are equal to BC, and Ap, pB are equal to Cp, pB. Now (Eucl. i. 20.) BC is less than Bp, pc, and therefore AP, PB are less than Ap, pB; therefore, &c. (7.) Of all straight lines which can be drawn from a given point to an indefinite straight line, that which is nearer to the perpendicular is less than the more remote. And from the same point there cannot be drawn more than two straight lines equal to each other, viz. one on each side of the perpendicular. Let A be the given point, and BC the given indefinite straight line. From A let fall the perpendicular AD, and draw any other lines AF, AG, AH, BED &c. of which AF is nearer to AD than AG is, and |