It is manifest (S. 6. 3.) that the point, so determined, is the point which was to be found. Otherwise. Let BCD and EFG be the two given circles, and A a given point without them: It is required to describe a circle which shall pass through A, and touch the two circles BCD, EFG. Find (E. 1. 3.) the centres, K and L, of the two given circles; draw KL and let it, produced, meet the circumference of BCD in B, and the circumference of EFG in E and G; and let it meet CH, which is drawn (S. 52. 3.) so as to touch both the circles, in H; join H, A; find (E.12.6.) a fourth proportional to AH, HB, and HG, and from HA cut off HI equal to it; so that (E. 16. 6.) BH X HG = AHX HI; lastly, describe (S. 95. 3.) a circle AMI, passing through A and I, and touching either of the given circles BCD, in H 1 some point, M; it shall, also, if HM be drawn, pass through the point N, in which HM cuts the circumference of the circle EFG, and shall touch EFG in the point N. For, if it be possible, let the circumference of the circle AMI cut HM in some other point, as P: Then (E. 36. 3.) MH × HP = AH × HI; but (constr.) AH × HI = BH × HG, and (E. 36. 3.) BHXHG=MHxHN; ... MHX HP = MHX HN; ... HP is equal to HN, the less to the greater, which is absurd; ... the circumference of the circle MIA cannot but meet the circle EFG in the point where it is cut by MH; and it touches the circle EFG in that point. For draw KM, KC, KD, LQ, LF, and LP, and let MK and PL, produced, meet in R: Then since (constr. and E. 18. 3.) the HFL, HCK, having a common ∠ at H, have the / HFL, HCK, right, they are (S. 26. 1.) equiangular; ... (E. 4. 6.) HL: LF or LQ:: HK: KC or KM: And the HLQ, HKM, have a common & at H, and have the two remaining & HQL, HΜΚ of the same species; for since MH cuts both the circles, the HQL, HMK, are (E. 16.3. cor.) each of them less than a right 2; .. (E. 7. 6.) the 2 HQL = ∠ HMK; .. (E. 15. def. 1. and E. 5. 1.) the ∠ LNQ, or RNM, = ∠ HMK, or NMR; ...(E. 6. 1.) RM = RN; but, since the circle AMI touches the circle BCD, of which K is the centre, ... (E. 11. or 12. 3.) the centre of AMI must be in MR; and since RM=RN, that centre (E. 7. 3.) must be in R; since, ..., the diameters of the two circles MAI, EFG, have a common extremity at N, the two circles (S. 6. 3.) touch one another.* PROP. LXI. 70. PROBLEM. To describe a circle that shall touch three given circles. Find a point (S. 59. 6.) such that the difference between its distances from the centres of the first and second of the given circles, shall be equal to the difference of the diameters of those circles, and such that the difference between its distances from the centres of the first and third of the given circles, shall be equal to the difference of the diameters of those circles: Then it is manifest, that the difference between its distances from the centres of the second and third of the given circles, will be equal to the difference of their diameters; and that, if from the point so determined, as a centre, a circle be described touching any one of the given circles, it will (S. 6. 3.) also touch the other two. * It is evident that Prop. 59 may be deduced from this proposition, as it is thus independently demonstrated; and that the proposition immediately following, which is one of some celebrity, may be deduced from either of them. PROP. LXII. 71. PROBLEM. Upon a given finite straight line, to describe an equilateral and equiangular figure, having the number of its sides equal to four, eight, sixteen, &c.; or to three, six, twelve, &c.; or to five, ten, twenty, &c.; or to fifteen, thirty, sixty, &c. sides. In any circle inscribe (S. 14. 4. cor. 3.) an equilateral and equiangular rectilineal figure of any number of sides that is specified in the proposition; then upon the given finite straight line describe (E. 18. 6.) a rectilineal figure similar to it, and the problem will have been solved. PROP. LXIII. 72. THEOREM. Similar triangles, and similar polygons, are to one another as any rectilineal figure described upon any side of the one, is to a similar rectilineal figure similarly described upon the homologous side of the other. For (E. 20. 6.) the two given figures, and two similar figures thus similarly described, will have to one another the same duplicate ratio of that which the homologous sides have. PROP. LXIV. 73. PROBLEM. To cut off from a given triangle any part required, by a straight line drawn parallel to a given straight line. Let ABC be the given A, and AX a given straight line: It is required to cut off, from the ABC, any assigned part, by a straight line drawn parallel to AX. First, let AX be parallel to BC; find (S. 21.6.) a square which shall be the same part of the square of AB, that the A, to be cut off, is required to be of the given A, and make AD equal to its side; through D draw (E. 31. 1.) DE parallel to AX or BC: Then is ADE the A which was to be cut off from ABC. 1 |