(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 finite 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.

=

D

H

Let AB be the given line, and 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; His 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.

с

in C. Take BD: AC in the given ratio, and from B draw BE parallel and equal to AC. Join DE and produce it to meet CF drawn at any angle from C, equal

G

B

to the given line; draw FG parallel to EB, and from G draw GH parallel to FC; G and Hare the points required. For BE being parallel to GF,

DG : GF :: ᎠᏴ : ᏴᎬ, or DG: HC :: DB: AC,

.. (Eucl. v. 19. Cor.)

BG: AH :: DB: AC in the given ratio,
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 C let any line ECG be drawn meeting them in E and

B

G, and join DE, DG; then DE: DG :: CE: CG.

For because AC: CB :: AD: DB
and EA: BG:: AC: CB,

(by sim. tri. ACB, BCG)

.. (Eucl. v. 11.) EA : BG:: DA : DB,

.. (Eucl. vi. 6.) the triangles EAD, GDB are equiangular, and ED: DG :: AE : BG :: CE: CG.

(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 EA 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

H

E

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

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, BA respectively; and take EF: EA in the given ratio. Divide

H

B

M

N

L

D G

K FO

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 BA 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 10 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,

.. IA: HB in the given ratio.

(31.) From two given straight lines to cut off two parts, which may have a given ratio; so that the ratio of the remaining parts may also be equal to the ratio of two other given lines.

E

Let AB be one of the given G lines; draw BG to make any angle with AB, and let BD be equal to the other given line. Take AB : BE in the given ratio of the remaining parts, and BF: BE in the given ratio of the parts to be cut off. and draw DH and BC parallel to EF, to DB meeting BC in C, and AB in I.

Join AE, FE, and HC parallel

Then (Eucl. vi. 2.) AI: IH :: AB: BE in the given ratio of the remainders; and the triangles BCI, BFE having the angle CBI= the alternate angle BFE, and CIB=FBE, are equiangular,

.. BI: IC :: BF : BE,

in the ratio of the parts to be cut off; and AB, HC (=DB) are the given lines.

(32.) Three lines being given in position, to determine a point in one of them; from which if two lines be drawn at given angles to the other two, the two lines so drawn may together be equal to a given line.

Let AB, AC, BC be the three lines given in position, take AD=the given line and making with AB an angle equal to one of the given angles. Through D draw Dba parallel to AB, and meeting AC and BC in a and b. Draw AE to meet CB in E making the angle AEC= the given angle to be made by the line to be drawn, with BC. In AE take Ad=AD, and join ad cutting BC in F. Draw FG parallel to EA meeting AC in G, which is the point required.