fore, two angles of the triangle BGC are 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. Because the angle BAC is equal to the angle ADB, each of them being a right angle, and the angle at B com mon to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD (4. Cor. 32. 1.): therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals (4. 6.); wherefore the triangles are similar (def. 1. 6.). In like manner, it may be demonstrated, that the triangle ADC is equiangulai and similar to the triangle ABC: and the triangles ABD, ADC, being each equiangular and similar to ABC, and equiangular and similar to one another. B D C 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 :: triangles ABC, BDA, BC: BA :: triangles ABC, ACD, BC CA DA: DC (4. 6.); and in the 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. Because ED is parallel to one of the sides of the A triangle ABC, viz. to BC, CD: D tion (18. 5), CA : AD :: BA : Al fore (C. 5.) BA is the same multip ber of times that AC contains AD AC, AE is the same of AB; wher required is cut off. PROP. 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 (Prop. 31. 1.) DF, EG, parallel to BC; and through D draw DHK, parallel to AB; therefore each of the figures FH, HB, is a parallelogram: wherefore DH is equal (34. 1.) to FG, and HK to GB: and because HE is parallel F to KC, one of the sides of the triangle DKC, CE: ED: (2. 6.) KH: HD; But KH=BG, and HDGF; therefore CE: ED:: BG: GF; Again, because FD is parallel to EG, one of the sides of the triangle AGE, ED: DA :: GF: FA; But it has been proved that CE :ED: BG: GF; therefore the given straight line AB is divided similarly to AC. PROP. XI. PROB. A D H E K G B 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 (Prop. 31.1.) Because BC is parallel to DE, a side of the triangle ADE, AB : (2. 6.) BD.: AC: CE; but BD=AC: therefore AB: AC:: AC CE. Wherefore to the two given straight lines AB, AC a third proportional, CF is found. A B C D E 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 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 (Prop. 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:: C: HF. Wherefore to the three given straight lines, A, B, C, a fourth proportional HF is found. 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. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B (Prop. 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 proportional between AB, BC, the seg A D B ments of the base (Cor 8. 6.); therefore hetween the two given straight lines AB, BC, a mean proportional DB is found. PROP. XIV. PROB. 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. 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, AB: FE::DB: BE (1. 6.), also, BC: FE:: GB: BF (1. 6.); therefore DB: BE:: GB: BF (11. 5.). 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 proportional, 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. 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 proportional, 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. 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 (14. 1.); join BD. Because the triangle ABC is equal to the triangle ADE, and ABD is an other 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, the triangles ABC, EAD have the same ratio to the triangle BAD; wherefore the triangle ABC is equal (9. 5.) to the triangle EAD. PROP. XVI. THEOR. If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means; And if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals. Let the four straight lines, AB, CD, E, F, be proportionals, viz. as AB to CD, so E to F; the rectangle contained by AB, F is equal to the rect angle contained by CD, E. E From the points A, C draw (11. 1.) AG, CH at right angles to AB, CD; and make AG equal to F, and CH equal to E, and complete the parallelograms BG, DH. Because AB: CD:: E: F; and since E-CH, and F=AG, AB : CD (7. 5.) : : CH: AG; therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another (14. 6.); therefore the parallelogram BG is equal to the parallelogram DH: and the parallelogram BG is contained by the straight lines AB, F; because AG is equal to F; and the parallelogram DH is contained by CD and E, because CH is equal to E: therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. F G A H And if the rectangle contained by B C D these four lines are proportionals, viz. AB is to CD as E to F. the straight lines AB, F be equal to that which is contained by CD, E; |