Next, let each of the angles at C, Fbe not less than a right angle. Then the triangle ABC shall also in this case be equiangular to the triangle DEF. G E F The same construction being made, it may be proved in like manner that BC is equal to BG, and therefore the angle at C'equal to the angle BGC: but the angle at Cis not less than a right angle; (hyp.) therefore the angle BGC is not less than a right angle: wherefore two angles of the triangle BGC are together not less than two right angles: which is impossible; (1. 17.) and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle: in this case likewise the triangle ABC shall be equiangular to the triangle DEF. For, if they be not equiangular, at the point B in the straight line AB make the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC: and therefore the angle BCG equal to the angle BGC: (1. 5.) but the angle BCG is a right angle, (hyp.) therefore the angle BGC is also a right angle; (ax. 1.) whence two of the angles of the triangle BGC are together not less than two right angles; which is impossible: (I. 17.) therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q. E.D. PROPOSITION VIII. THEOREM. In a right-angled triangle, if a perpendicular be drawn from the rightangle to the base; the triangles on each side of it are similar to the whole triangle, and to one another. Let ABC be a right angled-triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC. Then the triangles ABD, ADC shall be similar to the whole triangle ABC, and to one another. Because the angle BAC is equal to the angle ADB, each of them being a right angle, (ax. 11.) and that the angle at B is common to the two triangles ABC, ABD: the remaining angle ACB is equal to the remaining angle BAD; (1..32.) therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals; (VI. 4.) wherefore the triangles are similar: (VI. def. 1.) 1 in the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC. And the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right-angled, &c. Q. F. D. COR. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base; and also that each of the sides is a mean proportional between the base, and the segment of it adjacent to that side: because in the triangles BDA, ADŬ; BD is to DA, as DA to DC; (VI. 4.) and in the triangles ABC, DBA; BC is to BA, as BA to BD: (vI.4.) and in the triangles ABC, ACD; BC is to CA, as CA to CD. (VI.4.) PROPOSITION IX. PROBLEM. From a given straight line to cut off any part required. Let AB be the given straight line. It is required to cut off any part from it. E 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 parallel to CB. Then AE shall be the part required to be cut off. Because ED is parallel to BC, one of the sides of the triangle ABC, as CD is to DA, so is BE to EA; (VI. 2.) and by composition, CA is to AD, as BA to AE: (v. 18.) but CA is a multiple of AD; (constr.) therefore BA is the same multiple of AE: (v. D.) whatever part therefore AD is of AC, AE is the same part of AB: wherefore, from the straight line AB the part required is cut off. Q. E. F. PROPOSITION X. PROBLEM. 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 divided given straight line have. Let AB be the straight line given to be divided, and AC the divided line. It is required to divide AB similarly to AC. Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, E draw DF, EG parallels to BC. (1. 31.) Then AB shall be divided in the points F, G, similarly to ÀC. Through D draw DHK parallel to AB: therefore each of the figures, FH, HB is a parallelogram; wherefore DH is equal to FG, and HK to GB: (1.34.) and because HE is parallel to KC, one of the sides of the triangle ᎠᏦᏟ, as CE to ED, so is KH to HD: (vI. 2.) but KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF: (v. 7.) again, because FD is parallel to GE, one of the sides of the triangle AGE, as ED is to DA, so is GF to FA: (vi. 2.) therefore, as has been proved, as CE is to ED, so is BG to GF, and as ED is to DA, so is GF to FA: therefore the given straight line AB is divided similarly to AC. Q.E.F. PROPOSITION XI. PROBLEM. To find a third proportional to two given straight lines. It is required to find a third proportional to AB, AC. Let AB, AC be placed so as to contain any angle produce AB, AC to the points D, E; and make BD equal to AC; join BC, and through D, draw DE parallel to BC. (1. 31.) but BD is equal to AC; therefore as AB is to AC, so is AC to CE. (v. 7.) Wherefore, to the two given straight lines AB, AC, a third proportional CE is found. Q. E. F. PROPOSITION XII. PROBLEM. To find a fourth proportional to three given straight lines. It is required to find a fourth proportional to A, B, C.. Take two straight lines DE, DF, containing any angle EDF: and upon these make DG equal to A, GE equal to B, and DH equal to C; (I. 3.) join GH, and through E draw EF parallel to it. (1. 31.) but DG is equal to A, GE to B, and DH to C; Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Q. E. F. PROPOSITION XIII. PROBLEM. To find a mean proportional between two given straight lines, It is required to find a mean proportional between them. Piace AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw BD at right angles to AC. (1. 11.) Then BD shall be a mean proportional between AB and BC. Join AD, DC. And because the angle ADC in a semicircle is a right angle, (II. 31.) and because in the right-angled triangle ADC, BD is drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the base: (VI. 8. Cor.) therefore between the two given straight lines AB, BC, a mean proportional DB is found. Q.E.F. PROPOSITION XIV. THEOREM. 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 conversely, parallelograms that 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. 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 reciprocally proportional; that is, DB shall be to BE, as GB to BF. Let the sides DB, BE be placed in the same straight line; And because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB is to FE, as BC to FE: (v. 7.) but as AB to FE so is the base DB to BE, (vI. 1.) and as BC to FE, so is the base GB to BF; therefore, as DB to BE, so is GB to BF. (v. 11.) Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional. Next, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF: the parallelogram AB shall be equal to the parallelogram BC. and as DB to BE, so is the parallelogram AB to the parallelogram FE; (VI. 1.) and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FE: (v. 11.) therefore the parallelogram AB is equal to the parallelogram BC. (v. 9.) Therefore equal parallelograms, &c. Q.E.D. |