distances, and any point without the perpendicular is at unequal distances from the extremities of the line. From C the point of bisection let CD be drawn at right angles to AB; any point D is at equal distances from A and B. A 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. (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 C B E D P 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. (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 с E B 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. Let A and B be the given points and DE the line given in position. From Alet fall the perpendicular AD, and produce it to C making DC = AD. Join CB, AP. AP, PB will be the lines required. = 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 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 E 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 manif pC 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, &c. of which AF is nearer to I E D FGH -C B¬ AG than AH; then AF will be less than AG and AG than AH. For since the angle at D is a right angle, the angle AFG is greater than a right angle (Eucl. i. 16.), and therefore greater than AGF, hence (Eucl. i. 19.) AG is greater than AF. In the same manner it may be shewn that AH is greater than AG. = And from A there can only be drawn to BC two straight lines equal to each other, viz. one on each side of AD. Make DE=DF and join AE. Then AEAF (i. 2.). And besides AE no other line can be drawn equal to AF. For, if possible, let AI-AF. Then because AI= AF and AF=AE, therefore AI=AE, i. e. a line more remote is equal to one nearer the perpendicular, which is impossible; therefore AI is not equal to AF. In the same manner it may be shewn that no other but AE can be equal to AF, therefore, &c. (8.) Through a given point, to draw a straight line, so that the parts of it intercepted between that point and perpendiculars drawn from two other given points may have a given ratio. Let A and B be the points from which the perpendiculars are to be drawn, and C the point through which the line is to be drawn. Join AC, and produce it to D, making AC: CD in · the given ratio; join BD, and through C draw ECF perpendicular to BD. ECF is the line required. Draw AE parallel to BD, and .. perpendicular to EF. The triangles ACE, DFC, having each a right angle, and the angles at C equal, are equiangular, whence CE: CF:: AC: CD, i, e. in the given ratio. (9.) From a given point between two indefinite right lines given in position, to draw a line which shall be terminated by the given lines, and bisected in the given point. Let AB, AC be the given lines, meeting in A. From P the given point draw PD parallel to AC one of the lines, and make DE DA. Join EP and produce it to F, then will EF be bisected in P.. = A B E D P FC For since DP is parallel to AF, (Eucl. vi. 2.) EP: PF:: ED: DA, i. e. in a ratio of equality. COR. If it be required to draw a line through P which shall be terminated by the given lines, and divided in any given ratio in P, draw PD parallel to AC and take AD: DE in the given ratio and draw EPF, it will be the line required. (10.) From a given point without two indefinite right lines given in position; to draw a line such that the parts intercepted by the point and the lines may have a given ratio. Let AB, AC be the given lines, and P the given point. Draw PD parallel to AC, and take AD: DE in the given ratio. Join PE and produce it to F. Then PF: PE will be in the given ratio. For the triangles PDE and AEF are similar, having the angles at E equal, as also the angles PDE, EAF, (Eucl. i. 39.) P E B |