cess of each segment above that distance PD. And since GP is parallel to AB,

GP: PK :: AD: DB :: GH: (HP =) Pl,

hence (Eucl. v. 19. Cor.) GH: (PI=) PH :: PH : IK, and .. GH : IK in the duplicate ratio of GH : HP, i. e. of GP: PK.

(26.) If perpendiculars be drawn to two sides of a triangle from any two points therein; the distance of their concourse from that of the two sides will be to the distance between the two points, as either side is to the perpendicular drawn from its extremity upon the other.

From any two points E, F in the sides AB, AC of the triangle ABC, let perpendiculars ED, FD be drawn, meeting in D. Join AD, EF, and from C draw CG perpendicular to AB; AD : FE AC: CG.

H

EG

Produce ED to H. And since the angles AED, AFD are right angles, a circle described on AD as a diameter will pass through F and E, and .. the angles FAD, FED standing in the same segment are equal; .. the triangles AHD, HEF are equiangular;

and .. AD: FE :: AH: HE:: AC: CG,

since HE is parallel to CG.

(27.) If the three sides of a triangle be bisected, the perpendiculars drawn to the sides at the three points of bisection, will meet in the same point.

Let the sides of the triangle ABC be bisected in the points D, E, F. Draw the perpendiculars EG, FG meeting in G. The perpendicular at D also passes through G.

=

D

F

B

Join GD, GA, GB, GC. Since AF FC, and FG is common to the triangles AFC, CFG, and the angles at F are right angles, .. AG= GC. In the same way it may be shewn that GC GB; .. AG=GB; but AD = DB, and DG is common to the triangles ADG, BDG, .. the angles at D are equal and .. right angles, or the perpendicular at D passes through G.

COR. The point of intersection of the perpendiculars is equally distant from the three angles.

(28.) If from the three angles of a triangle lines be drawn to the points of bisection of the opposite sides, these lines intersect each other in the same point.

Let the sides of the triangle ABC be

GF; BGF is a straight line.

Join EF, meeting CD in H. Then

H

(Eucl. vi. 2.) FE is parallel to AB, and... the triangles DAG, GEH are equiangular,

.. DA DG :: HE: HG,

or ᎠᏴ : DG :: HF : HG,

i.e. the sides about the equal angles are proportional; ..the triangles BDG, GHF are similar, and the angle DGB=HGF; and .. BG and GF are in the same straight line.

(29.) The three straight lines, which bisect the three angles of a triangle, meet in the same point.

Let the angles BAC, BCA be bisect

ed by the lines AE, CD, and through G their point of intersection draw BGF; it bisects the angle at A.

D

E

F

For (Eucl. vi. 3.) BC : CF :: BG: GF :: BA : AF, .. BC: BA:: CF: FA,

or FB bisects the angle ABC.

(30.) If the three angles of a triangle be bisected, and one of the bisecting lines be produced to the opposite side; the angle contained by this line produced, and one of the others is equal to the angle contained by the third, and a perpendicular drawn from the common point of intersection of the three lines to the aforesaid side.

Let the three angles of the triangle ABC be bisected by the lines AD, BD, CD; produce BD to E, and from D draw DF perpendicular to AC; the angle ADE is equal to CDF.

ET

B

Since the three angles of the triangle ABC are equal to two right angles, .. the angles DAB, DBA, DCF are together equal to one right angle, i. e. to DCF, and CDF; whence the two angles DAB, DBA are together equal to the angle CDF; but ADE is equal to the same two angles, and .. ADE is equal to CDF.

(31.) In a right-angled triangle, if a straight line be drawn parallel to the hypothenuse, and cutting the

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

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 equal 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

:

and ABC BPC:: L: PD;

.. 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 LPD+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

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

K

G