to BO; that is, AO is to OQ as QO is to OB. Therefore the triangles AOQ and QOB are similar triangles (VI. 6); and therefore AQ is to QB as QO is to BO. This shews that the ratio of AQ to BQ is constant; we have still to shew that this ratio is the same as the assigned ratio.

We have already shewn that AO is to DO as DO is to BO; therefore, the difference of AO and DO is to Do as the difference of DO and BO is to BO (V. 17); that is, AD is to DO as BD is to BO; therefore AD is to BD as DO is to BO; that is, AD is to DB as QO is to BO. This shews that the ratio of QO to BO is the same as the assigned ratio.

ON MODERN GEOMETRY.

56. We have hitherto restricted ourselves to Euclid's Elements, and propositions which can be demonstrated by strict adherence to Euclid's methods. In modern times various other methods have been introduced, and have led to numerous and important results. These methods may be called semi-geometrical, as they are not confined within the limits of the ancient pure geometry; in fact the power of the modern methods is obtained chiefly by combining arithmetic and algebra with geometry. The student who desires to cultivate this part of mathematics may consult Townsend's Chapters on the Modern Geometry of the Point, Line, and Circle.

We will give as specimens some important theorems, taken from what is called the theory of transversals.

Any line, straight or curved, which cuts a system of other lines is called a transversal; in the examples which we shall give, the lines will be straight lines, and the system will consist of three straight lines forming a triangle.

We will give a brief enunciation of the theorem which we are about to prove, for the sake of assisting the memory in retaining the result; but the enunciation will not be fully comprehended until the demonstration is completed.

57. If a straight line cut the sides, or the sides produced, of a triangle, the product of three segments in order is equal to the product of the other three segments.

Let ABC be a triangle, and let a straight line be drawn cutting the side BC at D, the side CA at E, and the side AB produced through B ́at F. Then BD and DC are

called segments of the side BC, and CE and EA are called segments of the side CA, and also AF and FB are called segments of the side AB.

Through A draw a straight line parallel to BC, meeting DF produced at H.

Then the triangles CED and EAH are equiangular to one another; therefore AH is to CD as AE is to EC (VI. 4). Therefore the rectangle AH, EC is equal to the rectangle CD, AE (VI. 16).

Again, the triangles FAH and FBD are equiangular to one another; therefore AH is to BD as FA is to FB (VI. 4). Therefore the rectangle AH, FB is equal to the rectangle BD, FA (VI. 16).

Now suppose the straight lines represented by numbers in the manner explained in the notes to the second Book of the Elements. We have then two results which we can express arithmetically: namely, the product AH. EC is equal to the product CD.AE; and the product AH.FB is equal to the product BD.FA.

Therefore, by the principles of arithmetic, the product AH.EC.BD.FA is equal to the product AH.FB.CD.AE, and therefore, by the principles of arithmetic, the product BD.CE.AF is equal to the product DC.EA.FB.

This is the result intended by the enunciation given above. Each product is made by three segments, one from

every side of the triangle: and the two segments which terminated at any angular point of the triangle are never in the same product. Thus if we begin one product with the segment BD, the other segment of the side BC, namely DC, occurs in the other product; then the segment CE occurs in the first product, so that the two segments CD and CE, which terminate at C, do not occur in the same product; and so on.

The student should for exercise draw another figure for the case in which the transversal meets all the sides produced, and obtain the same result.

58. Conversely, it may be shewn by an indirect proof that if the product BD.CE. AF be equal to the product DC.EA.FB, the three points D, E, F lie in the same straight line.

59. If three straight lines be drawn through the angular points of a triangle to the opposite sides, and meet at the same point, the product of three segments in order is equal to the product of the other three segments.

Let ABC be a triangle. From the angular points to the opposite sides let the straight lines AOD, BÕE, COF be drawn, which meet at the point 0: the product AF.BD.CE shall be equal to the product FB. DC. EA.

For the triangle ABD is cut by the transversal FOC, and therefore by the theorem in 57 the following products are equal, AF.BC.DO, and FB.CD.OA.

Again, the triangle ACD is cut by the transversal EOB, and therefore by the theorem in 57 the following products are equal, AO.DB.CE and OD. BC. EA.

Therefore, by the principles of arithmetic, the following products are equal, AF. BC. DO. AO. DB. CE and FB.CD.OA.OD. BC. EA. Therefore the following

E

F

products are equal, AF. BD. CE and FB.DC.EA. We have supposed the point O to be within the triangle: if O be without the triangle two of the points D, E, F will fall on the sides produced.

60. Conversely, it may be shewn by an indirect proof that if the product AF. BD. CE be equal to the product FB. DC. EA, the three straight lines AD, BE, CF meet at the same point.

61. We may remark that in geometrical problems the following terms sometimes occur, used in the same sense as in arithmetic; namely arithmetical progression, geometrical progression, and harmonical progression. A proposition respecting harmonical progression, which deserves notice, will now be given.

62. Let ABC be a triangle; let the angle A be bisected by a straight line which meets BC at D, and let the exterior angle at A be bisected by a straight line which meets BC, produced through C, at E: then BD, BC, BE shall be in harmonical progression.

B

A

For BD is to DC as BA is to AC (VI. 3); and BE is to EC as BA is to AC (VI. A). Therefore BD is to DC as BE is to EC (V. 11). Therefore BD is to BE as DC is to EC (V. 16). Thus of the three straight lines BD, BC, BE, the first is to the third as the excess of the second over the first is to the excess of the third over the second. Therefore BD, BC, BE are in harmonical progression.

This result is sometimes expressed by saying that BE is divided harmonically at D and C.