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(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 AC; therefore AC is the least.
(2.) If a perpendicular be drawn bisecting a given straight line; any point in this perpendicular is at equal
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.
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 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,
Join AE, EB. Since AF=FB,
(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,
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 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 p C be joined, it may be shewn that Ap=p C. Hence AP and BP together are equal to BC, and Ap, p B are equal to Cp, p B. 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 sidle 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, B&c. of which AF is nearer to AD than AG is, and
F G H
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 AE=
= AF (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