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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 A 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.
ing in A.
(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, und bisected in the given point. Let AB, AC be the given lines, meet
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.
For since DP is parallel to AF, (Eucl. vi. 2.)
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.)
.:. FE: EP :: AE: ED and comp. PF: PE :: AD : DE, i. e. in the given ratio.
(11.) From a given point to draw a straight line, which shall cut off from lines containing a given angle, segments that shall have a given ratio.
Let ABC be the given angle, and P the given point, either without or within. In BA take any point A, and take AB : BC in the given ratio. Join AC, and from P draw PDE parallel to AC. PDE is the line required.
For since DE is parallel to AC, (Eucl. vi. 2.) DB : BE :: AB : BC, i. e. in the given ratio.
(12.) If from a given point any number of straight lines be drawn to a straight line given in position, to determine the locus of the points of section which divide them in a given ratio.
Let A be the given point, and BC the line given in position. From A draw any line AB, and divide it at E in the given ratio ; through E draw EF parallel to BD; it is the locus required.
From A draw any other line AD meeting EF in F; then (Eucl. vi. 2.) AF: FD :: AE : EB, i. e. in the given ratio. In the same manner any other line drawn from A to BD will be divided in the given ratio by EF which therefore is the locus required.
(13.) A straight line being drawn parallel to one of the lines containing a given angle and produced to meet the other ; through a given point within the angle, to draw a line cutting the other three, so that the part intercepted between the two parallel lines may have a given ratio to the part intercepted between the given point and the other line.
Let ABC be the given angle, DE parallel to AB, and P the given point.
From P draw PC parallel to DE or AB, and take BE: CF in the given ratio. Join FP and produce it to A; APF is the line required. For since DE and CP are parallel to AB,
AD: EB :: DF: EF :: PF: CF :. alt. AD: PF:: EB: CF i. e. in the given ratio.
(14.) Two parallel lines being given in position ; to draw a third, such that, if from any point in it lines be drawn at given angles to the parallel lines, the intercepted parts may have a given ratio.
Let AB, CD be the given parallel lines; in AB take any point E; and draw EF, EG making angles equal to the given angles; produce EF, and take EH: EG in the given ratio ; produce FE to I so that FI: IE :: HE: EF; through I draw LI parallel to AB, it is the line required.
Draw IK parallel to EG ; then the triangles IEK, EFG are equiangular,
.. IE: IK :: EF: EG
but FI: IE :: HE: EF; ... ex æqu. FI : IK :: HE: EG, i. e. in the given ratio; and IKE=EGF which is one of the given angles, and by construction IFG is equal to the other. Also lines drawn from any point in LI, making with AB and CD angles equal to the given angles will be parallel and equal to FI, IK, and :: in the given ratio.
(15.) If three straight lines drawn from the same point and in the same direction be in continued proportion, and from that point also a line equal to the mean proportional be inclined at any angle; the lines joining the extremity of this line and of the proportionals will contain equal angles.
Let AB : AC :: AC: AD, and from A let AE be drawn equal to AC, inclined at any angle to AB; join EB, EC, ED; the angle BEC is equal to the angle CED. For since AB : AC :: AC : AD, and AE= AC;
.. AB : AE :: AE : AD, i.e. the sides about the angle A are proportional, and :: the triangles AEB, AED are similar, and the angle AED is equal to EBA. Also since CA=AE, the angle AEC=ECA; but ECA is equal to the two BEC, EBC, (Eucl. i. 32.) .. also AEC is equal to the two BEC, EBC; of which DEA is equal to EBC; .:. the remainder DEC is equal to the remainder BEC.