therefore also in the duplicate ratio of AE ED; whence from the definition of the duplicate ratio, AE ED :: ED: EC.

(23.) To draw a line parallel to the common base of two triangles which have different altitudes, so that the parts of it intercepted by the sides may have a given ratio.

Let ABC, DBC. be two triangles on the same base BC, the vertex D being in the side AC. Divide BC in E, so that BC: CE may be equal to the given ratio. Join AE, cutting BD in G; and through G draw FH parallel to BC; FH is the line required.

Since FH is parallel to BC, FH : GH: BC: CE, i. e. in the given ratio.

F

G

LH

E

But if the vertex I is not in AC, draw ID parallel to BC; join BD; divide the base BC, as before; join AE, and draw FK parallel to BC. Then it is evident that GH=LK, and .. FH: LK in the given ratio.

COR. If the triangles be upon equal bases, but in the same straight line, the line may be drawn in a similar

manner.

(24.) If the base of a triangle be produced, so that the whole may be to the part produced in the duplicate ratio of the sides; the line joining the vertex and the extremity of the part produced will be a mean proportional between the whole line produced and the part produced.

Let AC be produced to D, so that AD may be to DC in the duplicate ratio of AB: BC; join BD; it will be a mean proportional between AD and DC.

A

B

C

Draw CE parallel to AB; then AB : CE :: AD : DC, i. e. in the duplicate ratio of AB BC, whence AB: BC:: BC: CE, i. e. the sides about the equal angles ABC, BCE are proportional; therefore the triangles ABC, BCE are similar, and the angle at A is equal to the angle CBD; .. the triangles ABD, CBD are equiangular, and

AB: BD :: BD : DC.

(25.) To determine a point within a given triangle, which will divide a line parallel to the base into two segments, such that the excess of each segment above the perpendicular distance between the parallel lines may to each other in the duplicate ratio of the respective segments.

Let ABC be the given triangle. From C draw CD perpendicular to AB, and from D draw DE, DF bisecting the angles ADC, BDC. Join BE, cutting CD in P; P is the point required.

E

H

be

F

PL

K

Through P draw GHIK parallel to AB; then the angle PDH is equal to the angle HDA, i. e. to the alternate angle PHD; and .. HP, and in like manner PI will each be equal to PD the perpendicular distance of GK from AB; and GH, IK will be equal to the ex

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

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

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.

F

H

Produce ED to H. And since the angles AED, AFD are right angles, a circle described on AD as a diameter will pass through Fand 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.

=

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 bisected in D, E, F. Join AE, CD, meeting each other in G. Join BG, GF; BGF is a straight line.

G

H

Join EF, meeting CD in H. Then (Eucl. vi. 2.) FE is parallel to AB, and.. the triangles DAG, GEH are equiangular,

.. DA DG HE: HG,

:

or DB: 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.

A

B

E

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