From the point A draw a straight line AC, making any angle with AB; and 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 draw DE pa- B rallel to it then AE is the part required. C For, because ED is parallel to BC, one of the sides of the triangle ABC, therefore BE is to EA as CD to DA, and, by composition (5. 18), BA is to AE as CA to AD: But CA is a certain multiple of AD; therefore (5. D), BA is the same multiple of AE, that is, whatever part AD is of AC, AE is the same part of AB: Wherefore, from the given straight line AB, the part required has been cut off. Q. E. F. PROP. X. PROB. To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another, which the parts of the given divided line have. Let AB be the straight line given to be divided, and AC the given divided line; it is required to divide AB similarly to AC. F H Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle, A and join BC; through the points D, E draw DF, EG parallel to BC, and through D draw DHK parallel to AB: Then each of the figures FH, HB is a parallelogram, and therefore DH is equal to FG, and HK to GB: Now, because HE is parallel to KC, one of the sides of the triangle DKC, therefore KH is to HD as CE to ED: But KH is equal to BG, and HD to GF; therefore BG B K is to GF as CE to ED: Again, because FD is parallel to GE, one of the sides of the triangle AGE, GF is to FA as ED to DA: And it has been proved that BG is to GF as CE to ED, and GF to FA as ED to DA; therefore the given straight line AB is divided similarly to AC. Q.E.F. PROP. XI. PROB. To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle: it is required to find a third proportional to AB, AC. Produce AB, AC, to the points D, E, and make BD equal to AC; join BC, and through D draw DE parallel to BC: Then, because BC is parallel to DE, a side of the triangle ADE, therefore AB is to BD as AC to CE: But BD is equal to AC; B therefore AB is to AC as AC to CE: Wherefore, to the two given straight lines AB, AC, D a third proportional CE has been found. Q. E. F. D A H F PROP. XII. PROB. To find a fourth proportional to three given straight lines. it is Let A, B, C, be three given straight lines required to find a fourth proportional to A, B, C. Take two straight lines, DE, DF, containing any angle EDF, and in these make DG equal to A, GE equal to B, and DH equal to C; join GH, and through E draw EF parallel & to GH: Then, because GH is parallel to EF, one of the sides B A B C C E of the triangle DEF, therefore DG is to GE as DH to HF: But DG is equal to A, GE to B, and DH to C; therefore A is to B as C to HF: Wherefore, to the three given straight lines A, B, C, a fourth proportional HF has been found. Q.E. F. PROP. XIII. PROB. To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines: it is required to find a mean proportional D between them. Place AB, BC in a straight line, and B C upon AC describe the semicircle ADC; A from the point B draw BD at right angles to AC, and join AD, DC: Then the angle ADC in a semicircle is a right angle; and because in the right-angled triangle ADC, DB is drawn from the right angle perpendicular to the base, therefore (6. 8. Cor.) DB is a mean proportional between AB, BC, the segments of the base: Wherefore, between the two given straight lines AB, BC, a mean proportional DB has been found. Q. E. F. PROP. XIV. THEOR. Equal parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and parallelograms, which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. First, let AB, BC be equal parallelograms, which have the angles at B equal: the sides of the parallelograms AB, BC about the equal angles shall be recipro cally proportional, that is, DB shall be to BE as GB to BF. Let the parallelograms be placed, so that the sides DB, BE may be in the same straight line; then, since the angles DBF, EBF are together equal to two right angles, and that the angle DBF is equal to the angle GBE, therefore the angles GBE, EBF are together equal to two right angles, and therefore GB, BF are in the same straight line: Complete the parallelogram FE: Then, because the parallelogram AB is equal to BC, and that FE is another parallelogram, therefore AB is to FE as BC to FE: But as AB is to FE as the base DB to the base BE, and BC is to FE as A the base GB to the base BF; therefore DB is to BE as GB to BF. D F B E G с Next, let AB, BC be parallelograms, which have the angles at B equal and the sides about the equal angles reciprocally proportional, viz. DB to BE as GB to BF: the parallelogram AB shall be equal to the parallelogram BC. For, the same construction being made, because DB is to BE as GB to BF, and that DB is to BE as the parallelogram AB to the parallelogram FE, and GB to BF as the parallelogram BC to the parallelogram FE, therefore AB is to FE as BC to FE, and therefore (5.9) the parallelogram AB is equal to the parallelogram BC. Wherefore, Equal parallelograms &c. Q. E.D. PROP. XV. THEOR. Equal triangles, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and triangles, which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. First, let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE: the sides of the triangles about the equal angles shall be reciprocally proportional, that is, CA shall be to AD as EA to AB. Let the triangles be placed so that their sides CA, AD may be in the same straight line, and therefore also EA and AB will be in the same straight line; join BD: Then, because the triangle ABC is equal to the triangle ADE, and that ABD is another triangle, therefore the triangle ABC is to the triangle ABD as the triangle ADE to the triangle ABD: But the triangle ABC is to the triangle ABD as the base CA to the base o B AD, and the triangle ADE is to the triangle ABD as the base EA to the base AB; therefore CA is to AD as EA to AB. Next, let ABC, ADE be triangles, which have the angle BAC equal to the angle DAE, and the sides about the equal angles reciprocally proportional, viz. CA to AD as EA to AB: the triangle ABC shall be equal to the triangle ADE. For, the same construction being made, because CA is to AD as EA to AB, and CA is to AD as the triangle ABC to the triangle ABD, and EA to AB as the triangle ADE to the triangle ABD, therefore the triangle ABC is to the triangle ABD as the triangle ADE to the triangle ABD, and therefore (5. 9) the triangle ABC is equal to the triangle ADE. Wherefore, Equal triangles &c. Q. E.D. |