Let ABC, DEF be two triangles having the angle BAC equal to EDF, and ABC equal to the exterior angle DFG, made by producing the side EF;

then AC CB :: DE: EF.

At the point D in the line FD, make the angle FDG equal to the angle EDF or BAC, and meeting EF pro

duced in G. Since the angle FDG is equal to the angle BAC, and DFG is equal to ABC, .. the triangles ABC, DFG are equiangular, and

AC CB: DG : GF,

But since the angle GDE is bisected by DF, .. (Eucl. vi. 3.)

DG: GF:: DE : EF,

.. AC CB :: DE: EF.

:

(22.) In a given triangle to draw a line parallel to one of the sides, so that it may be a mean proportional between the segments of the base.

Let ABC be the given triangle; in the base of which take a point E, such that AE may be to EC in the duplicate ratio of AC: CB; draw ED parallel to BC; ED is the line required.

E C

Since ED is parallel to BC, AE : ED :: AC : CB. But AE EC in the duplicate ratio of AC: CB, and

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.

A

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.

B

I

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.

Draw CE parallel to AB; then AB DC, i. e. in the duplicate ratio of AB

B

E

CE: AD:
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 be 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.

F

E

K

G

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

excess 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.

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 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.

=

F

B

Join GD, GA, GB, GC. Since AF FC, and FG is common to the triangles AFG, 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

F

B

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.