Hence the two sets of conditions for the similarity of AEFH, ABCD are satisfied (Art. 91). In like manner FKCG is similar to ABCD. .. AEFH, FKCG are similar. [Prop. 25. Art. 174. PROPOSITION LIV. (Euc. VI. 26.) ENUNCIATION. If two similar parallelograms have a common angle and be similarly situated they are about the same diagonal. Let ABCD, AEFH be two similar parallelograms having the same angle A. Let them be similarly situated, and let AB, AE be corresponding sides. It is required to prove that the diagonals AC, AF coincide in direction. Since the parallelograms are similar ABC = AÊF, AB: AE BC: EF H A E B [Prop. 28. Hence the triangles ABC, AEF are similar. In these triangles BC, EF are corresponding sides. Hence the angles opposite them are equal. :. BẬC = EÂF. Hence AC coincides in direction with AF. Hence the parallelograms ABCD, AEFH are about the same diagonal. Art. 175. EXAMPLE 80. Let the straight line AB be produced through A to P and through B to Q, so that AP is equal to BQ. On BQ, BP let similar parallelograms be similarly described, viz. BQRS and BPTU. Prove that the parallelogram whose adjacent sides are QA, QR is equal to that whose adjacent sides are PA, PT. Art. 176. PROPOSITION LV. (Euc. VI. 27, 28, 29.) ENUNCIATION. If OAB be a given triangle it is required to find a point P on AB or AB produced so that if PQ be drawn parallel to OB to cut OA in Q, and if PR be drawn parallel to OA to cut OB in R, then the parallelogram PQOR may have a given area. There are two kinds of cases. Case I. Suppose the point P to have been found and to lie between A and F, the middle point of AB. Hence if P be between A and F the area OQPR is less than the triangle OAF. (This is equivalent to the result of Euc. VI. 27.) = (given triangle OAB) – (a given area). Hence the parallelogram SPTF has a known area. It is also known to be similar to the known parallelogram EAVF. Hence it can be constructed by Prop. 47, and if it be placed so that the side corresponding to FV falls along FV, and the side corresponding to FE falls along FE, then its diagonal will fall on FA by Prop. 54. Hence the position of P is known. (This is equivalent to the result of Euc. VI. 28.) In order that the construction for P may be possible it is necessary that the given area should not exceed half the given triangle OAB. The above construction applies only to the case where P lies between A and F, the middle point of AB. H. E. 14 If P be one position of the required point, let a point P' be taken on FB so that PF = P'F, and let Q', R', S', T' be the points corresponding to Q, R, S, T. Then the parallelograms SPTF, S'P'T'F are equal. Hence the parallelograms OQPR, OQ'P'R' are equal. Hence P' is another position of the required point. Case II. Let P be on BA produced through A, and let the same construction be made. Then OQPR = OESR + EQPS = EAUS+ PUVT =SPTF-EAVF SPTF ΔΟΑΡ. .. SPTF▲OAF+ OQPR = (given triangle OAB) + (a given area). Hence the parallelogram SPTF has a known area. It is also known to be similar to the known parallelogram EAVF. Hence it can be constructed by Prop. 47, and if it be placed so that the side corresponding to FV falls along FV, and the side corresponding to FE falls along FE, then its diagonal will fall along FA by Prop. 54. Hence the position of P is known. (This is equivalent to the result of Euc. VI. 29.) In this case the construction is always possible for all magnitudes of the given area. If P be one position of the required point, and a point P' be taken on FB produced through B so that P'F= PF, then it may be shown that P' is another position of the required point. It results from Cases I and II that if the given area be less than half the given triangle OAB there are four solutions of the problem, viz. P may be between A and F or between F and B, or on BA produced through A, or on AB produced through B. If the given area be equal to half the triangle OAB there are three solutions, viz. P may be at F, or on BA produced through A, or on AB produced through B. If the given area be greater than half the triangle OAB there are two solutions, viz. P may be on BA produced through A, or on AB produced through B. SECTION X. THE REMAINING IMPORTANT PROPOSITIONS IN THE THEORY OF SCALES AND OF RATIOS, WITH GEOMETRICAL APPLICATIONS. Props. 56-63. Art. 177. PROPOSITION LVI.* (Euc. V. 23.) ENUNCIATION 1. If A, B, C are three magnitudes of the same kind, if if and prove that if ENUNCIATION 2. A: B=U: V, B:C=TU, ACT: V. If A, B, C are three magnitudes of the same kind, As in Props. 22 and 35 it is best to make use of the second form of the conditions in Prop. 8 and it is necessary to show that (1) If rA <sC, then rT < sV, (2) If rA > sC, then rT> s V, (3) If rT <sV, then rA < sC, * See Note 11. |