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

E

D

to the given line; draw FG parallel to EB, and from G draw GH parallel to FC; G and H are 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,

(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

E

B

H

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

AC

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

H

D G

N

B

M

K FO E

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

D

B

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.

H

F

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.

For through G draw IGH parallel to DA, then the triangles aGI, a AD are similar, and

and.. GI= GF, .. GH+GF= GH+GI= AD=the given line, and the angle GHB = DAB and GFC= AEC, . GHB, GFC are equal to the given angles.

(33.) If from a given point two straight lines be drawn including a given angle, and having a given ratio, and one of them be always terminated by a straight line, given in position, to determine the locus of the extremity of the other.

Let A be the given point, and BC the line given in position. From A draw any line AD, and make the angle DAE equal to the given angle, and take AE such that AD: AE may be in the given ratio, and through E draw EF making the angle AEF-ADB; EF is the locus required.