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:. FC : DH :: CE : ED :: 2:1, :. FC=2 DH; but since AI, BG, DH are parallel, and AD= DB, . . AI+BG=2 DH=FC.
(25.) From a given point in one of two straight lines given in position, to draw a line to cut the other, so that if from the point of intersection a perpendicular be let fall upon the former, the segment intercepted between it and the given point, together with the first drawn line may
be equal to a given line. Let AB, BC be the lines given in position, and A the
given point. Draw AD perpendicular to AB, and meeting BC in D; draw DE parallel to AB, and equal to the given line. And draw EF parallel to AD, meeting CB in F. Join FA, and produce it, and from D draw DG=DE, meeting FG in G, and draw AH parallel to DG, and let fall the perpendicular HI; AH and AI together are equal to the given line.
Through H draw KL parallel to DE; then since GD is parallel to AH and HL to DE; ..DG : AH :: FD : FH :: DE : HL,
but DG=DE, ... AH=HL,
(26.) One of the lines which contain a given angle, is also given. To determine a point in it such, that if from thence to the indefinite line there be drawn a line having a given ratio to that segment of it which is adjacent to the given angle; the line so drawn, and the other segment of the given line, may together be equal to another given line.
Let AB be the given line, and K BAC the given angle. From B draw BD to AC, such that it may be to AB in the given ratio*; produce it till BE=the other given line. Through E draw EC parallel to AB, meeting AC in C. Join BC, and draw DF so that it may=DE, and draw BG, GH respectively parallel to FD, EB; H is the point required.
For produce HG to meet CE in K; Then (Eucl. vi. 2.) ED : KG :: CD: CG :: DF: BG,
but ED=DF, ... KG= BG, and HG+GB=HG+GK=BE=the given line, and HG : HA :: BD : AB i. e. in the given ratio.
(27.) Two straight lines and a point in each being given in position; to determine the position of another point in each, so that the straight line joining these latter points may be equal to a given line, and their respective distances from the former points in a given ratio.
Let A and B be the given points in the lines AC, BD which are given in position, and produced to meet
* That is, the given ratio must be less than that of AB to the perpendicular on AD.
in C. Take BD : AC in the
DG : GF :: DB : BE,
and HG=CF= the given line.
(28.) If a straight line be divided into any two parts,
( and produced so that the segments may have the same ratio that the whole line produced has to the part produced, and from the extremities of the given line perpendiculars be erected; then any line drawn through the point of section, meeting these perpendiculars, will be divided at that point into parts, which have the same ratio that those lines have, which are drawn from the extremity of the produced line to the points of intersection with the perpendiculars.
Let AB be divided into any two parts in C and produced to D so that AC: CB :: AD: DB, and from A and B let AE, BF be drawn perpendiculars to AB, and through Clet any
line ECG be drawn meeting them in E and G, and join DE, DG; then DE: DG :: CE: CG.
For because AC : CB :: AD : DB
and EA : BG :: AC : CB, (by sim. tri. ACE, BCG) :: (Eucl. v. 11.) EA : BG :: DA : DB, ::: (Eucl. vi. 6.) the triangles EAD, GDB are equiangular,
and ED : DG :: AE : BG :: CE : CG.
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(29.) From two given points, to draw two straight lines which shall contain a given angle, and meet two lines given in position, so that the parts intercepted between those points and the lines may have a given ratio.
Let AB, CD be the lines given in position, and E, F the given points. From E draw E A perpendicular to AB, and make the angle AGF equal to the given angle. In GF produced take FH such, that the ratio of EA : FH may be the same as the given ratio. Draw HD perpendicular to GH meeting CD in D. Draw DFI and BEI to include the given angle. These are the lines required.
For, since the angles FGE, FIE are equal, as also FKG, EKI, .. GFK, IEK or their vertically opposite. angles DFH, AEB are equal, and the angles at H and A are right angles, .. the triangles FDH, AEB are equiangular, and
EB : FD :: EA : FH, i.e. in the given ratio.
(30.) The length of one of two lines which contain
a given angle being given; to draw from a given point without them a straight line which shall cut the given line produced, so that the part produced may be in a given ratio to the part cut off from the indefinite line.
Let AB be the given c line, and ABC the given angle; and D the given point. Draw AE, DE parallel to BC,B A respectively; and take EF: EA in the given ratio. Divide DF so that FE : DG :: FG : AB. Join AG; and draw DH parallel to AG, and it will be the line cutting BC in H, and B A produced in I, as was required.
Join AF; and draw BK parallel to AG cutting AF in L; and draw LM parallel to KE cutting AE in M and AG in N.
Then FE : LM :: GF: (NL=) AB
and FE : DG :: FG : AB by construction; ..LM=DG=IA; if therefore ILO be drawn, IL must be equal and parallel to AM, and IO to AE (Eucl. i. 33.). In the same manner it is evident that HB=IL= AM; and by similar triangles AFE, ALM,
FE : EA :: LM : MA
:: IA : HB
(31.) From two given straight lines to cut off two parts, which may have a given ratio; so that the ratio
a of the remaining parts may also be equal to the ratio of two other given lines.