.. these triangles are mutually equiangular. I. 32, Cor. 1

1. An equilateral triangle is inscribed in a circle, and from any point on the Oce straight lines are drawn to the vertices; prove that one of these is equal to the sum of the other two. 2. In all quadrilaterals that cannot be inscribed in a circle, the rectangle contained by the diagonals is less than the sum of the two rectangles contained by the opposite sides. 3. Prove the converse of the proposition.

4. ABC is a triangle inscribed in a circle; D, E are taken on

AB, AC so that B, D, E, C are concyclic; the circle ADE cuts the former in F. Prove that FE + FB: FC + FD= AB: AC. (R. Tucker.)

APPENDIX VI.

TRANSVERSALS.

DEF. 1.-When a straight line intersects a system of straight lines, it is called a transversal.

This definition of a transversal is not the most general (that is, comprehensive) one, but it will suffice for our present purpose.

PROPOSITION 1.

If a transversal cut the sides, or the sides produced, of a triangle, the product of three alternate segments taken cyclically is equal to the product of the other three, and conversely.*

G

B

E

B

G

Let ABC be a triangle, and let a transversal cut BC, CA, AB, or these sides produced at D, E, F respectively :

it is required to prove AF. BD. CE = FB. DC. EA.

Draw AG || BC, and meeting the transversal at G. Then As AFG, BFD are mutually equiangular; ... AF: AG = BF: BD;

I. 31

I. 29

VI. 4

VI. 16

* Given in the third book of the Spherics of Menelaus, who lived at Alexandria towards the close of the first century A.D. For a full account of the theorem, see Chasles' Aperçu Historique sur l'origine el le développement des Méthodes en Géométrie, p. 291.

Again, ▲s AEG, CED are mutually equiangular;

E

I. 29

VI. 4

VI. 16

Multiply equations (1) and (2) together, and strike out the common factor AG; then AF. BD. CE = FB. DC · EA.

COR. 1.-The equation AF BD · CE = FB · DC. EA may be put in any of the following four useful forms:

COR. 2.-Consider ABC as the triangle, DEF as the transversal;

then AFBD · CE = FB · DC. EA. (1)

Consider AFE as the triangle, BCD as the transversal;

then AB FD. EC = BF.DE CA. (2)

Consider BDF as the triangle, AEC as the transversal ; then BC. DE. FA

=

CD.EFAB. (3)

Consider CED as the triangle, AFB as the transversal; then CBDF · EA = BD.FE. AC. (4)

Any one of these four equations may be deduced from the other three by multiplying them together and striking out the factors common to both sides.

The converse of the theorem (which may be proved indirectly) is, If two points be taken in the sides of a triangle, and a third point in the third side produced, or if three points be taken in the three sides produced of a triangle, such that the product of three alternate segments taken cyclically is equal to the product of the other three, the three points are collinear.

PROPOSITION 2.

If three concurrent straight lines be drawn from the vertices of a triangle to meet the opposite sides, or two of those sides produced, the product of three alternate segments of the sides taken cyclically is equal to the product of the other three; and conversely.*

Let ABC be a triangle, and let AD, BE, CF, which pass through any point O, meet the opposite sides in D, E, F:

it is required to prove AF. BD. CE = FB. DC. EA.

Consider ABD as a triangle cut by the transversal COF; then AFBC · DO = FB · CD.OA. (1)

App. VI. 1

Consider ADC as a triangle cut by the transversal BOE;

then AO. DB. CE = OD · BC.EA. (2)

App. VI. 1

Multiply equations (1) and (2) together, and strike out the common factors AO, DO, BC;

then AF. BD. CE = FB · DC · EA.

COR.-Repeat Cor. 1 to the preceding theorem.

The converse of the theorem (which may be proved indirectly) is, If three straight lines be drawn from the vertices of a triangle to meet the opposite sides, or two of those sides produced, so that the product of three alternate segments of the sides taken cyclically is equal to the product of the other three, the three straight lines are concurrent.

* This theorem is first found in a work of the Marquis Giovanni Ceva, De lineis rectis se invicem secantibus, statica constructio (1678), Book I., Prop. 10. The proof given in the text is due to Carnot, the founder of the Theory of Transversals. See his Essai sur la Théorie des Transversales (1806), p. 74.

NOTE. To distinguish readily between the converse of Menelaus's theorem and that of Ceva's, it should be observed that in the first case an even number of the points D, E, F are situated on the sides, and an odd number on the sides produced; in the second case matters are reversed.

PROPOSITION 3.

If two triangles be situated so that the straight lines joining corresponding vertices are concurrent, the points of intersection of corresponding sides are collinear; and conversely.*

Let ABC, A'B'C' be two triangles such that AA', BB', CC' are concurrent at 0; and let the corresponding sides BC, B'C' meet in L, AC, A'C' in M, AB, A'B' in N:

it is required to prove L, M, N collinear.

Consider AOB as a triangle cut by the transversal A'B'N;

then AN BB' · OA' = NB · B'O · A'A. (1)

App. VI. 1

Consider AOC as a triangle cut by the transversal A'C'M;

then AA'. OC'. CM = A'O.C'C. MA. (2)

App. VI. 1

Consider BOC as a triangle cut by the transversal B'C'L; then B'OC'C. LB = BB'. OC'. CL. (3)

App. VI. 1

* Due to Girard Desargues, an architect of Lyon, who was born 1593, and died 1662. See Poudra's Euvres de Desargues, tome i. pp. 413, 430.