from another givent without them a line be drawn intersecting the others so as to be harmonically divided; then will all lines drawn from that point meeting the three lines be harmonically divided.

cally divided in C and B; then will any other line EF be harmonically divided in G and H. Through G draw IK parallel to BD,

then DC CB:: KG: GI

But DC: CB:: DE: EB

.. (Eucl. v. 15.) DE: EB :: KG

GI

and alt. DE : KG :: EB : GI and since DE is parallel to GK, (Eucl. vi. 2.) DE: KG :: EF : FG

and EB being parallel to GI,

.. EB GI :: EH : HG,

whence (Eucl. v. 15.)

EF: FG :: EH : HG

and alt. EF : EH :: FG : HG.

(23.) If a straight line be divided into two equal, and also into two unequal parts, and be produced, so that the part produced may have to the whole line so produced the same ratio that the unequal segments of the line have to each other; then shall the distances of the point of unequal section from one extremity of the given line,

from its middle point, from the remity of the part produced, and from the other extremity of the given line, be proportionals.

BD

A

Let AB be divided into two equal E parts in Cand into two unequal parts in D, and produced to E, so that BE : EA :: BD : DA; then will AD :

COR. The converse may easily be proved to be true.

(24.) Three points being given; to determine another, through which if any straight line be drawn, perpendiculars upon it from two of the former, shall together be equal to the perpendicular from the third.

Let A, B, C be the three given points. Join AB, and bisect it in D. Join CD, from which cut off DE equal to a third part of it. E is the point required.

F

E

H

B

Through E, let any line FG be drawn, and let fall on it the perpendiculars AI, BG, DH, CF; then the angles at F and H being right angles, and the vertical angles at E equal, the triangles CFE, DHE are equiangular,

.. FC: DH :: CE: ED :: 2 : 1, ..FC=2 DH; but since AI, BG, DH are parallel, and AD=DB, .. AI+BG=2DH = 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

ing BC in D;

given point. Draw AD perpendicular to AB, and meetdraw 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,

.. AH + AI= KL= DE = the given line.

(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

A

H

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.

C

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

B

D

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,

·.. (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 Clet any line ECG be drawn meeting them in E and

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