BOOK VI. DEFINITIONS. E 1. Two rectilineal figures are said to be equiangular to one another when the angles of the first, taken in order, are equal respectively to those of the second, taken in order. 2. Rectilineal figures are said to be similar when they are equiangular to one another, and also have the sides about the equal angles taken in order proportionals. Thus the two quadrilaterals ABCD, EFGH are similar if the angles at A, B, C, D are respectively equal to those at E, F, G, H, and if the following proportions hold : AB : BC :: EF : FG, с H G DA: AB :: HE: EF. In these proportions, sides which are both antecedents or both consequents of the ratios are said to be homologous or corresponding. [Def. 6, p. 320.] Thus AB and EF are homologous sides ; so are BC and FG. 3. Two similar rectilineal figures are said to be similarly situated with respect to two of their sides when these sides are homologous. 4. Two figures are said to have their sides about one angle in each reciprocally proportional when a side of the first figure is to a side of the second as the remaining side of the second figure is to the remaining side of the first. 5. A straight line is said to be divided in extreme and mean ratio when the whole is to the greater segment as the greater segment is to the less. ; PROPOSITION 1. THEOREM. [EUCLID'S PROOF.] [ The areas of triangles of the same altitude are to one another as their bases. Let ABC, ACD be two triangles of the same altitude, namely the perpendicular from A to BD. Then shall the A ABC : the A ACD :: BC : CD. Produce BD both ways; and from CB produced cut off any number of parts BG, GH, each equal to BC; and from CD produced cut off any number of parts DK, KL, LM, each equal to CD. Join AH, AG, AK, AL, AM. Since the A8 ABC, ABG, AGH are of the same altitude, and stand on the equal bases CB, BG, GH, .. the AS ABC, ABG, AGH are equal in area ; I. 38. ::. the A AHC is the same multiple of the A ABC that HC is of BC. Similarly the A ACM is the same multiple of the A ACD that CM is of CD. And if HC=CM, I. 38. and if HC is less than CM, the A AHC is less than the A ACM. I. 38, Cor. Now since there are four magnitudes, namely, the A' ABC, ACD, and the bases BC, CD; and of the antecedents, any equimultiples have been taken, namely, the A AHC and the base HC; and of the consequents, any equimultiples have been taken, namely the A ACM and the base CM; and since it has been shewn that the A AHC is greater than, equal to, or less than the A ACM, according as HC is greater than, equal to, or less than CM; .:: the four original magnitudes are proportionals; v. Def. 5. that is, the A ABC : the A ACD :: the base BC: the base CD. Q.E.D. COROLLARY. The areas of parallelograms of the same altitude are to one another as their bases. Let EC, CF be parms of the same altitude. Join BA, AD. I. 34. and the par CF is double of the A ACD; .. the parm EC : the par" CF :: BC : CD. NOTE. This proof of Proposition 1 is founded on Euclid's Test of Proportion, and therefore holds good whether the bases BC, CD are commensurable or otherwise. The numerical treatment given on the following page applies in strict theory only to the former case ; but the beginner would do well to accept it, at any rate provisionally, and thus postpone to a later reading the acknowledged difficulty of Euclid's Theory of Proportion. PROPOSITION 1. [NUMERICAL ILLUSTRATION.) The areas of triangles of equal altitude are to one another as their base3. А = B L M N C E R S F , and therefore of equal altitude. Then shall the ABC : the A DEF=the base BC : the base EF. Suppose BC contains 4 units of length, and EF 3 units; and let BL, LM, MN, NC each represent one unit, as also ER, RS, SF. Then BC: EF=4:3. Join AL, AM, AN ; also DR, DS. Then the four AS ABL, ALM, AMN, ANC are all equal; for they stand on equal bases, and are of equal altitude. .:. the A ABC is four times the A ABL. Similarly, the A DEF is three times the A DER. But the AS ABL and DER are equal, for they are on equal bases BL, ER, and of equal altitude ; hence the A ABC : the A DEF=4:3 = BC: EF. This reasoning holds good however many units of length the bases BC, EF contain. Thus if BC=m units, and EF=n units, then, whatever whole numbers m and n represent, the A ABC: the A DEF=m:n BC: EF. EXERCISES ON PROPOSITION 1. ; 1. Two triangles of equal altitude stand on bases of 6:3 inches and 5.4 inches respectively; if the area of the first triangle is 125 square inches, find the area of the other. [109 sq. in.] 2. The areas of two triangles of equal altitude have the ratio 24:17; if the base of the first is 4.2 centimetres, find the base of the second to the nearest millimetre. [3:0 c.m.] 3. Two triangles lying between the same parallels have bases of 16.20 metres and 20.70 metres; find to the nearest square centimetre the area of the second triangle, if that of the first is 50-1204 sq. metres. [60-0427 sq. m. ] 4. Assuming that the area of a triangle ={ base x altitude, prove algebraically that (i) Triangles of equal altitudes are proportional to their bases ; (ii) Triangles on equal bases are proportional to their altitudes. Also deduce the second of these propositions geometrically from the first. 5. Two triangular fields lie on opposite sides of a common base; and their altitudes with respect to it are 4.20 chains and 3.71 chains. If the first field contains 18 acres, find the acreage of the whole quadrilateral. [33:9 acres.] DEFINITION. Two straight lines are cut proportionally when the segments of one line are in the same ratio as the corresponding segments of the other. [See definition, page 139.] Fig. 1. Fig. 2. B х Thus AB and CD are cut proportionally at X and Y, if AX : XB :: CY : YD. And the same definition applies equally whether X and Y divide AB and CD internally as in Fig. 1 or externally as in Fig. 2. |