PROPOSITION 14. THEOREM. Parallelograms which are equal in area, and which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: Conversely, parallelograms which have one angle of the one equal to one angle of the other, and the sides about these angles reciprocally proportional, are equal in area. Let the parms AB, BC be of equal area, and have the - DBF equal to the GBE: then shall the sides about these equal angles be reciprocally proportional, that is, DB: BE :: GB: BF. Place the parms so that DB, BE may be in the same straight line; ... FB, BG are also in one straight line. Then because the parTM AB = the parTM BC, and FE is another parTM, 1. 14. Нур. .. the parTM AB: the partTM FE :: the parm BC: the parm FE; but the parTM AB: the parTM FE :: DB : BE, and the parm BC: the partTM FE :: GB : BF, ... DB: BE :: GB: BF. Conversely, let the DBF be equal to the and let DB: BE :: GB: BF. VI. 1. v. 1. GBE, Then shall the partTM AB be equal in area to the parm BC. For, with the same construction as before, by hypothesis DB : BE :: GB : BF; but DB: BE :: the partTM AB: the parTM FE, VI. 1. and GB: BF:: the partTM BC: the parm FE, .. the parm AB: the partTM FE:: the partTM BC: the partTM FE; v.1. ... the parm AB = the partTM BC. Q. E. D. PROPOSITION 15. THEOREM. Triangles which are equal in area, and which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: Conversely, triangles which have one angle of the one equal to one angle of the other, and the sides about these angles reciprocally proportional, are equal in area. S Let the ABC, ADE be of equal area, and have the ✓ CAB equal to the ∠EAD: then shall the sides of the triangles about these angles be reciprocally proportional, Place the that is, CA: AD :: EA : AB. $ so that CA and AD may be in the same st. line; ... BA, AE are also in one st. line. Join BD. Then because the ACAB = the EAD, and ABD is another triangle; 1. 14. Нур. Conversely, let the CAB be equal to the ∠EAD, and let CA: AD :: EA : ΑΒ. Then shall the CAB = AEAD. For, with the same construction as before, by hypothesis CA : AD :: EA : AB; ८ EXERCISES. ON PROPOSITIONS 14 AND 15. 1. Parallelograms which are equal in area and which have their sides reciprocally proportional, have their angles respectively equal. 2. Triangles which are equal in area, and which have the sides about a pair of angles reciprocally proportional, have those angles equal or supplementary. 3. AC, BD are the diagonals of a trapezium which intersect in 0; if the side AB is parallel to CD, use Prop. 15 to prove that the triangle AOD is equal to the triangle BOC. 4. From the extremities A, B of the hypotenuse of a rightangled triangle ABC lines AE, BD are drawn perpendicular to AB, and meeting BC and AC produced in E and D respectively: employ Prop. 15 to shew that the triangles ABC, ECD are equal in area. 5. On AB, AC, two sides of any triangle, squares are described externally to the triangle. If the squares are ABDE, ACFG, shew that the triangles DAG, FAE are equal in area. 6. ABCD is a parallelogram; from A and C any two parallel straight lines are drawn meeting DC and AB in E and F respectively; EG, which is parallel to the diagonal AC, meets AD in G: shew that the triangles DAF, GAB are equal in area. 7. Describe an isosceles triangle equal in area to a given triangle and having its vertical angle equal to one of the angles of the given triangle. 8. Prove that the equilateral triangle described on the hypotenuse of a right-angled triangle is equal to the sum of the equilateral triangles described on the sides containing the right angle. [Let ABC be the triangle right-angled at C; and let BXC, CYA, AZB be the equilateral triangles. Draw CD perpendicular to AB; and join DZ. Then shew hy Prop. 15 that the AYC=the ADAZ; and similarly that the BXC=the BDZ.] PROPOSITION 16. THEOREM. If four straight lines are proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means: Conversely, if the rectangle contained by the extremes is equal to the rectangle contained by the means, the four straight lines are proportional. Let the st. lines AB, CD, EF, GH be proportional, so that AB: CD :: EF: GH. Then shall the rect. AB, GH = the rect. CD, EF. From A draw AK perp. to AB, and equal to GH. 1. 11, 3. From C draw CL perp. to CD, and equal to EF. Complete the parms KB, LD. Then because AB: CD :: EF: GH; and EF = CL, and GH = AK; ..AB: CD :: CL: AK; Нур. Constr. that is, the sides about equal angles of parms KB, LD are reciprocally proportional; .'. KB = LD. But KB is the rect. AB, GH, for AK = GH, VI. 14. Constr. Conversely, let the rect. AB, GH = the rect. CD, EF: then shall AB: CD :: EF: GH. For, with the same construction as before, because the rect. AB, GH = the rect. CD, EF; Нур. Constr. that is, the parms KB, LD, which have the angle at A equal to the angle at C, are equal in area; ... the sides about the equal angles are reciprocally proportional: that is, AB: CD :: CL : AK; Q. E. D. PROPOSITION 17. THEOREM. If three straight lines are proportional the rectangle contained by the extremes is equal to the square on the mean: Conversely, if the rectangle contained by the extremes is equal to the square on the mean, the three straight lines are proportional. Let the three st. lines A, B, C be proportional, so that A:B::B: C. Then shall the rect. A, C be equal to the sq. on B. Take D equal to B. Then because A: B:: B: C, and D = B; ..A:B::D:C; VI. 16. .. the rect. A, C = the rect. B, D; ... the rect. A, C = the sq. on B. Conversely, let the rect. A, C = the sq. on B: For, with the same construction as before, Нур. and the sq. on B = the rect. B, D, for D = B; .. the rect. A, C = the rect. B, D, ..A:B:: D: C, that is, A: B :: B : C. VI. 16. Q. E. D. |