and because AB has the same ratio to each of the lines BC, BG ; BC is equal (9. 5.) to BG, and therefore the angle BGC is equal to the angle BCG (6.1): But the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle (13. 1.) But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle: But by the hypothesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal, that is, they are equal: And the angle at A is equal to the angle at D; wherefore the remaining angle at C is equal to the remaining angle at F: Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle; the triangle ABC is also, in this case, equiangular to the triangle DEF. The same construction being made, it may be proved, in like manner, that BC is equal to BG and the angle at C equal to the angle BGC: But the angle at C is not less less than a right angle; therefore the angle BGC is not less than a right angle: Wherefore, two angles of the triangle BGC are D 44 CE together not less than two right angles, which is impossible (17. 1.); and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. PROP. VIII. THEOR. In a right angled triangle if a perpendicular be drawn from the right angle 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 the triangles ABD, ADC are similar to the whole triangle ABC, and to one another. A Because the angle BAC is equal to the angle ADB, each of them being a right angle, and the angle at B common to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD (32. 1.): therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportion B D C als (4.6.); wherefore the triangles are similar (def. 1. 6.). In the like manner, it may be demonstrated, that the triangle ADC is equian : gular and similar to the triangle ABC and the triangles ABD, ADC, being each equiangular and similar to ABC, are equiangular and similar to one another. Therefore, in a right angled, &c. Q. E. 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 its segment adjacent to that side. For in the triangles BDA, ADC, BD DA: DA: DC (4 6.); and in the triangles ABC, BDA, BC : BA :: BA: BD (4. 6.); and in the triangles ABC, ACD, BC CA: CA: CD (4. 6.). PROP. IX. PROB. From a given straight line to cut off any part required, that is, a part which shall be contained in it a given number of times. Let AB be the given straight line; it is required to cut off from AB, a part which shall be contained in it a given number of times. From the point A draw a straight line AC making any angle with AB; and in AC take any point D, and take AC such that it shall contain AD, as oft as AB is to contain the part, which is to be cut off from it; join BC, and draw DE parallel to it: then AE is the part required to be cut off. E A D Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, CD: DA :: DE: EA (2. 6.); and by composition (18. 5.), CA: AD BA AF: But CA is a multiple of AD; therefore (C. 5.) BA is the same multiple of AE, or contains AE the same num ber of times that AC contains AD ; and there fore, whatever part AD is of AC, AE is the same of AB; wherefore, from the straight line AB the part required is cut off. Which was to be done. B 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 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 (31. 1.) DF, EG, parallel to BC; and through D draw DHK, parallel te AB; therefore each of the figures FH, HB, is a parallelo. 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; and having joined BC, through D draw DE parallel to it (31. 1.). A B C D E Because BC is parallel to DE, a side of the triangle ADE, AB: (2. 6.) BD:: AC: CE; but BE AC: therefore AB: AC AC: CE. Wherefore to the two given straight lines AB, AC a third proportional, CE is found. Which was to be done. PROP. XII. PROB. To find a fourth proportional to three given straight lines. Let A, B, C be the 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; and having joined GH, draw EF parallel (31. 1.) to it through the point E. And because GH is parallel to EF, one of the sides of the triangle DEF, DG: GE :: DH: HF (2. 6.); but DG=A, GE=B, and DH C; and therefore A: B:: CHF Wherefore to the three given straight lines, A, B, C, a fourth proportional HF is found. Which was to be done. 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 between them. D Place AB, BC in a straight line. and upon AC describe the semicircle ADC, and from the point B (11. 1.) draw BD at right angles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angle (31. 3.) and because in the right angled triangle ADC, DB is drawn from the right angle, perpendicular to the base, DB is a mean propor tional between AB, BC, the segments of the base (Cor. 8. 6.); therefore between the two given straight lines AB, BC, a mean proportional DB is found. Which was to be done. PROP. XIV. THEOR Equal parallelograms which have one angle of the one equa, 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. Let AB, BC be equal parallelograms, which have the angles at B equal, and let the sides DB, BE be placed in the same straight line; wherefore also FB, BG are in one straight line (14. 1.) : the sides of the parallelograms AB, BC, about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB to BF. Complete the parallelogram FE; and because the parallelograms AB, BC are equal, and FE is another parallelogram, AB: FE: BC: FE (7. 5.): but because the parallelograms AB, FE have the same altitude, BC FE :: GB : BF (1.6); therefore Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional. But, let the sides about the equal angles be reciprocally proportion al, viz. as DB to BE, so GB to BF; the parallelogram AB is equal to the parallelogram BC. : Because, DB BE :: GB: BF, and DB: BE :: AB: FE, and GB: BF: BC EF, therefore, AB FE :: BC: FE (11. 5.): Wherefore the parallelogram AB is equal (9. 5.) to the parallelogram BC. Therefore 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 in the one equal to one angle in the other, and their sides about the equal angles reciprocally pro portional, are equal to one another. Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE; the sides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA to AB. B A E Let the triangles be placed so that their sides CA, AD be in one straight line; wherefore also EA and AB are in one straight line (11. 1.); join BD. Because the triangle ABC is equal to the triangle ADE, and ABD is another triangle; therefore, triangle CAB : triangle BAD triangle EAD: triangle BAD; but CAB BAD :: CA: AD and EAD : BAD :: EA : AB; therefore CA: AD :: EA : AB (11. 5), wherefore the sides of the triangles ABC, ADE about the equal angles are reciprocally proportional. : But let the sides of the triangles ABC, ADE, about the equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the triangle ABC is equal to the triangle ADE. Having joined BD as before; because CA: AD: : EA: AB; and since CA: AD: : triangle ABC triangle BAD (1. 6.); and also EA: AB: triangle EAD: triangle BAD (11. 5.); therefore, triangle ABC triangle BAD: : triangle EAD: triangle BAD; that is, |