A E Draw a straight line AC making any angle with AB; in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it; join BC, and (i. 31.) draw DE parallel to it: AE is the part required. Because ED is parallel to BC, one of the sides of the triangle ABC, CD:DA::BE: EA (VI. 2.); and, by composition, (V. 18.) CA: AD::BA: A E. But (const.) CA is a multiple of AD; therefore (V. D.) BA is the same multiple of AE. Whatever part, therefore, AD is of AC, the same part is AE of AB. Therefore, from the straight line AB the part required is cut off : which was to be done. B с PROP. X. PROB. F H G E To divide a given straight line similarly to a given divided straight line; that is, into parts proportional to the parts of the given divided line. Let AB be the straight line given to be divided, and AC the given line divided in D and E; it is required to divide AB similarly to AC. Let AB, AC be placed so as to contain any angle; join BC, and, (1. 31.) through the points D, E, draw DË, EG parallels to it: AB is divided in F and G similarly to AC. Through D draw DHK parallel to AB. Then each of the figures FH, HB is (1. def. 24.) a parallelogram; wherefore DH is equal (I. 34.) to FG, and HK to GB. And (VI. 2.) because HE is parallel to KC, one of the sides of the triangle DKC, as CE:ED:: KH:HD; but KH is equal to BG, and HD to GF; therefore, as CE: ED:: :: BG:GF. Again, (VI. 2.) because FD is parallel to EG, as ED: DA::GF:FA. But it has been proved that CE:ED:: BG: GF; and now that ED:DA ::GF: FA : therefore the given straight line AB is divided similarly to AC: which was to be done. Schol. In a similar manner we may produce a given straight line so that the whole line so produced may have to the part produced the ratio of two given straight lines. Thus, if AB be the line to be produced, make at A an angle of any magnitude, and take AC and CD equal to the other given lines ; join BD, and draw CE parallel to it. Then, since EC is parallel to BD, a side of the triangle ABD, we have (VI. 2.) AE: EB:: AC:CD, so that AE has to EB the given ratio. B с А D B E с PROP. XI. PROB. To find a third proportional to two given straight lines. Let A and B be two given straight lines; it is required to find a third proportional to them. Take two straight lines CF, CG, containing any angle C; and upon these make CD equal to A, and DF, CE each equal to B. Join DE, and (I. 31.) draw FG parallel to it. EG is the third proportional required. For (VI. 2.) since DE is parallel to FG, CD: DF:: CE: EG. But (const.) CD is equal to A, and DF, CE each equal to B : therefore A:B::B: EG: wherefore to A and B, the third proportional EG is found, which was to be done. * с A B E G PROP. XII. PROB. To find a fourth proportional to three given straight lines. Let A, B, C be three given straight lines ; it is required to find a fourth proportional to them. Other modes of solving this problem may sometimes be employed with advantage. The following are among the most useful : 1. Draw AD (fig. to prop. 8.) perpendicular to the indefinite straight line BC, and make DB, DA equal to the given lines : join AB, and draw AC perpendicular to it: DC is the third proportional required. For (VI. 8. cor.) BD: DA::DA: DC. 2. Draw BC perpendicular to AB, and having made B A and AC equal to the less and greater of the given lines, draw CD and BE perpendicular AC: AD will be a third proportional to A B and AC, and AE to AC and AB. This follows from the fourth proposition of this book, since the triangles ABC, ACD are equiangular, as are also ABC, AEB. The angles ABC, ACD, &c., are here made right angles. They may be of any magnitude, however, provided they be equal. It is sufficient, therefore, to draw two straight lines, AB, AC, making any angle; to cut off AB, AC equal to the given proportionals ; and then, BC being joined, to make the angle ACD equal to ABC, and to draw BE parallel to CD; or to make the angle A BE equal to ACB, and to draw CD parallel to BE. This method affords an easy means of continuing a series of lines in continual proportion, both ways, when any two successive terms are given. Thus, after CD and BE are drawn, it is only necessary to draw DF, EG, &c., parallel to BC, and FH, GK, &c., parallel to CD; as AD, AF, AH, &c., will be the succeeding terms of the ascending series, and AE, AG, AK, &c., those of the descending one. GB D A B Take two straight lines DE, DF, containing any angle EDF, and make DG equal to A, GE* equal to B, and DH equal to C; and having joined GH, draw (1.31.) EF parallel to it through the point E: HF is the fourth proportional required. For, (VI. 2.) since GH is parallel to EF, as DG: GE:: DH:HF; but DG is equal to A, GE to B, and DH to C; therefore as A:B::C: HF: wherefore, to the three given straight lines, A, B, C, the fourth proportional HF is found : which was to be G E done. + PROP. XIII. PROB. To find a mean proportional between two given straight lines. Let AB, BC be two given straight lines; it is required to find the mean proportional between them. Place AB, BC in a straight line, and upon AC as diameter * It is plain that GE may be taken on either side of G. + The solution of this problem may also be effected in several different ways; some of which may be merely indicated to the student, as the proof's present no difficulty ; and it is evident that, with slight modification, they are applicable in solving the 1lth proposition, wbich is only a particular case of the 12th. 1. Let AE, EC (last fig. to III. 35., page 85) be the second and third terms, placed contiguous, and in the same straight line, and draw BE, making any angle with AC, and equal to the first term; through the three points, A, B, C, describe a circle cutting BE produced in D: ED is the fourth proportional. If AB and CD be joined, the proof will be obtained by means of the triangles ABE, CDE, which are similar. 2. Draw AB, AC (fig. to III. 36., cor, page 86) making any angle, and make AB, AE equal to the second and third terms: then if AC be taken equal to the first term, and a circle be described passing through B, C, and E, and meeting AC in F, AF will be the required line. If BF and EC be joined, the triangles A BF and ACE are similar, and hence the proof is immediately obtained. 3. Make BD and DA (fig. to VI. 8., page 145) perpendicular to each other, and equal to the first and second terms; join AB, and draw AC perpendicular to it; in DA, produced through A, if necessary, take a line equal to the third term, through the upper extremity of which draw a line parallel to AC: the line intercepted on DC, produced if necessary, between D and this parallel is the fourth proportional required. 4. If any number of antecedents, and the consequent of one of them be given, the other consequents may be readily found on the principle employed in the tenth proposition of this book. Thus, it' AD (fig. to that proposition, page 146) be the given consequent and AF its antecedent; then, if FG, GB, &c., be the other antecedents, the segments DE, EC, &c., made on the other line by the parallels, are the remaining consequents. describe the semicircle ADC; from B (I. 11.) draw BD at right angles to AC: BD is the mean proportional between AB and BC. Join AD, DC. Then, because the angle ADC in a semicircle is (111. 31.) a right angle, and because in the right-angled triangle ADC, DB is drawn from the right angle perpendicular to AC, DB is a mean proportional (VI. 8. cor.) between AB, BC the segments of the base. Therefore between AB, BC, the mean proportional DB is found : which was to be done.* B PROP. XIV. THEOR. EQUAL parallelograms which have an angle of the one equal to an angle of the other, have their sides about those angles reciprocally proportional: and (2.) parallelograms which have an angle of the one equal to an angle of the other, and the sides about those angles reciprocally proportional, are equal to one another. 1. Let AB, BC be equal parallelograms, which have the angles at B equal; the sides about those angles are reciprocally proportional; that is, DB : BE::GB : BF. Let the sides DB, BE be placed in the same straight line, and contiguous, and let the parallelograms be on opposite sides of DE; then, (I. 14. cor.) because the angles at B are equal, FB, BG are in one straight line. Complete the parallelogram FE, † and * Out of several additional ways of solving this problem, the following may be mentioned. 1. On the greater extreme as diameter describe a semicircle; from the diameter cut off a segment equal to the less extreme, through the extremity of which draw a perpendicular cutting the circumference ; and the chord drawn from that intersection to the extremity of the diameter common to the two extremes is the required mean. The proof is manifest from III. 31., and VI. 8. cor. 2. Make AD, DC (2d fig. to III. 36., page 86) equal to the given extremes ; on AC as chord describe any circle, and a tangent drawn from D will be the required mean. The proof of this is obtained by joining AB and BC, as the triangles ADB, BDC are similar. When one mean is determined, others may be found between it and the given extremes, and thus three means will be inserted between the given lines ; and by finding means between each successive pair of the five terms of which the series then consists, the number of means will be increased to seven. By continuing the process we may find fifteen means, thirty-one means, or any number which is less by one than a power of 2. We cannot find, however, by element. ary geometry, any other number of means, such as two, four, or five. + If AD, CG were produced to meet, it would be easy to show, that AB and BC would be the complements of the parallelograms about the diagonal of the whole parallelogram AC. In the demonstration it should in strictness be proved that AF and CE meet when produced. This follows from I. 29. cor. I. A F E B AB is equal to BC, and that FE is another parallelogram, AB:FE:: BC: FE. But (VI. 1.) as AB to FE, so is the base DB to BE; and as BC to FE, so is the base GB to BF; therefore (V. 11.) as DB:BE::GB: BF. The sides, therefore, of the parallelograms AB, BC, about their equal angles, are (VI. def. 2.) reciprocally proportional. 2. But let the sides about the equal angles be reciprocally proportional, viz., DB : BE:: GB: BF; the parallelograms AB, BC are equal. The same construction being made, because, as DB : BE:: GB : BF; and (VI. 1.) as DB : BE :: AB: FE; and as GB: BF:: BC: FE; therefore (V. 11.) as AB: FE:: BC: FE: wherefore (V. 9.) the parallelogram AB is equal to the parallelogram BC. Therefore equal parallelograms, &c. PROP. XV. THEOR. EQUAL triangles which have an angle of the one equal to an angle of the other, have their sides about those angles reciprocally proportional : and (2.) triangles which have an angle of the one equal to an angle of the other, and the sides about those angles reciprocally proportional, are equal to one another. 1. Let ABC, ADE be equal triangles, which have the angles BAC, DAE equal; the sides about those angles are reciprocally proportional; that is, CA: AD :: EA : AB. Let CA, AD be placed in one straight line, and contiguous, and let the triangles be on opposite sides of CD; then (I. 14. cor.) BA, AE are in one straight line. Join BD : and because the triangles ABC, ADE are equal, and ABD is another triangle ; therefore (V. 7.) as CAB is to BAD, so is EAD to BAD. But(VI. 1.) as the triangle CAB to BAD, so is CA to AD; and as the triangle EAD to BAD, so is EA to AB; as, therefore, (V. 11.) CA : AD :: EA : AB ; wherefore (VI. def. 2.) the sides of the triangles ABC, ADE about the equal angles are reciprocally proportional. 2. But let the sides of the triangles ABC, ADE, about the equal angles be reciprocally proportional, viz., CA:AD :: EA : AB; the triangles ABC, ADE are equal. The same construction being made, because CA: AD:: EA: AB; and (VI. 1.) B D C E |