perpendicular drawn from the right angle; and through the point of intersection a line be drawn from one of the acute angles to the opposite side, and the extremity of this line and of the perpendicular be joined; the locus of its intersection with the line parallel to the hypothenuse will be a straight line.

Let EF be drawn parallel to AC the hypothenuse of the right-angled triangle ABC; and from the right angle B let the perpendicular DB be

E

H

K

B

F

drawn, meeting EFin G; through G draw CGH; join HD; the locus of I, the intersection of EF and HD is a straight line.

Because EG is parallel to AC the base of the triangles AHC, ABD, AK : KD :: EI: IG :: AD: DC. But AD and DC are invariable, .. the ratios of AK: KD, and EI: IG are also. In the same manner if any other line be drawn parallel to the hypothenuse, and a similar construction be made, the point of intersection will divide the part intercepted between AB and BD in the ratio of AD: DC, or AK: KD, and will .. be in the line BK, which is the locus required.

(32.) If from the angles of a triangle, lines, each équal to a given line, be drawn to the opposite sides (produced if necessary); and from any point within, lines be drawn parallel to these, and meeting the sides of the triangle; these lines shall together be equal to the given line.

From the angles of the triangle ABC let the lines. Aa, Bb, Cc be drawn to the opposites sides, each equal

to a given line L; and parallel to them respectively draw, from any point P, the lines PD, PE, PF; these together will be equal to L.

Join PA, PB, PC. Then since the triangles ABC, APC are on the same base AC, they are to one another

as the perpendiculars from B and P, i. e. by similar triangles, as Bb: PE, or as L: PE. In the same way, ABC ABP :: L: PF,

:. ABC : APC+ABP +BPC :: L : PE + PF+PD; and since the first term is equal to the second, the third will be equal to the fourth, or L=PD+PE+PF.

(33.) If the sides of a triangle be cut proportionally, and lines be drawn from the points of section to the opposite angles; the intersections of these lines will be in the same line, viz. that drawn from the vertex to the middle of the base.

F

D

Ο

Let the sides of the triangle ABC be cut proportionally, so that AD : AE :: DF EG :: FH: GL :: HB : LC. Join BE, BG, BL, CD, CF, CH; these lines will intersect each other in the line AK drawn from A to K the middle of the base BC.

H

B

Join DE. Then since when any number of magnitudes are proportional, as one antecedent is to its consequent, so are all the antecedents taken together to all

the

consequents together, .. AD : AE :: AB : AC, and DE is parallel to BC. Join KO, and let it meet DE in I. The triangles BOK, IOE are similar, and therefore,

BK: KO :: EI: 10, and for the same reason,

CK : KO :: DI: 10, whence EI=DI, and DE is bisected by KO; and it is also bisected by AK, :. AK passes through O. In the same manner it may be shewn that BG and CF, as also BI, CH intersect each other in points which are in the line AK.

(34.) If from any point in one side of a triangle, two lines be drawn, one to the opposite angle, and the other parallel to the base, and the former intersect a line drawn from the vertex bisecting the base; this point of intersection, that of the line parallel to the base and the third side, and the third angular point are in the same straight line.

From any point D in the side AB of the triangle ABC, let DE be drawn. parallel to AC, and DC joined; and let DC meet BF drawn from B to the middle of AC in G; A, G, E are in the same straight line.

D

B

E

G

Let DE cut BF in K. The triangles DGK, CGF are equiangular, and

.. DG GC :: DK : FC :: DE: AC;

hence the triangles DGE, AGC, having one angle in each equal, viz. EDG, GCA, and the sides about them proportional, are therefore similar; whence the angles AGC, DGE are equal; and DGC being a straight line, AGE is also.

(35.) If one side of a triangle be divided into any two parts, and from the point of section two straight lines be drawn parallel to, and terminating at the other sides, and the points of termination be joined; and any other line be drawn parallel to either of the two former lines, so as to intersect the other, and to terminate in the sides of the triangle; then the two extreme parts of the three segments into which the line so drawn is divided will always be in the ratio of the segments of the first divided line.

B

M

D

N

E

H

K

L

Let AB be divided into any two parts in D, from which draw DE, DF parallel to the other two sides of the triangle; join EF, and draw GH parallel to DE, meeting DF and EF in I and K; GI: KH :: AD: DB; and if LM be parallel to DF, LK MN; AD: DB.

:

Since GI is parallel to AF, and NK to DF,

GI : AF :: (ID=) NK : FD :: NE : DE :: KH : FC, .. GI : KH :: AF : FC :: AD : DB, since DF is parallel to BC. Again since ML is parallel to BC,

MN: BE :: ND : DE :: KF : FE :: KL : EC, .. MN: KL :: BE : EC :: BD: DA.

P

(36.) If through the point of bisection of the base of a triangle any line be drawn, intersecting one side of the triangle, and the other produced, and meeting a parallel to the base from the vertex; this line will be cut harmonically.

From the vertex B of the E triangle ABC, let BE be drawn parallel to the base AC, and through the middle point D let any line EGF be drawn meeting AB, BC, BE, in F,

F

A

FE FD BE : AD,

D

but BE (DC=) DA :: EG GD, since the triangles BGE, DGC are equiangular, .. (Eucl. v. 15.) EG: GD;: FE: FD,

or the line is divided harmonically.

(37.) If from either angle of a triangle a line be drawn intersecting that which joins the vertex and the bisection of the base, the opposite side, and the line from the vertex parallel to the base; it will be cut harmonically.

From the vertex A of the triangle ABC, let AE be drawn parallel to the base BC, and AD to its point of bisection D; and from C draw any line CFGE; then will CE CF:: EG: FG.

E

B

H

F