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draw FK parallel to AD or BC; therefore if each of the figures Book IV.
K the four straight lines GE, GF, GH, GK are equal to one another; and the circle described from the centre G, at the distance of one of them, shall pass through the extremities of the other B H с three, and touch the straight lines AB, BC, CD, DA; because the angles at the points E, F, H, K are right d angles, and that the straight line which is drawn from d 29. 1; the extremity of a diameter, at right angles to it, touches the circlee; therefore each of the straight lines AB, BC, CD, DA e 16. 3. touches the circle, which therefore is inscribed in the square ABCD. Which was to be done.'
PROP. IX. THEOR. ER 03.
To describe a circle about a given square.
Let ABCD be the given square ; it is required to describe a circle about it.
Join AC, BD cutting one another in E; and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the
A two BA, AC; and the base DC is equal
D to the base BC; wherefore the angle
E DAC is equal to the angle BAC, and
a 8. ko , the angle DAB is bisected by the straight, line AC: in the same manner, it may be demonstrated that the angles ABC, BCD, CDA are severally · bisected by the
C straight lines BD, AC; therefore, be, cause the angle DAB is equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA wherefore the side AE is equal to the side EB: in the same manner, it may be b 6. 1;
Book iv, demonstrated that the straight lines EC, ED are each of them
equal to EA or EB; therefore the four straight lines EA, EB, EC, ED are equal to one another; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD. Which was to be done.
PROP. X. PROB,
To describe an isosceles triangle, having each of the angles at the base double of the third angle,
*. 11. 2.
B without the
Take any straight line AB and divide a it in the point C, so that the rectangle AB, BC be equal to the square of. CA; and from the centre A, at the distance AB, describe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE ; join DA, DC, and about the triangle ADC describes the circle ACD; the triangle ABD-is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD.
Because the rectangle AB, BC is equal to the square of AC and that AC, is equal to BD, the rectangle AB, BC is equal to the square of BD; and because
'e 32. 3.
1 31. 1.
but BDA is equals to the angle CBD, because the side AD Book tv. is equal to the side AB; therefore CBD, or DBA is equal to war BED; and consequently the three angles BDA, DBA, BCD g 5.1. are equal to one another; ant because the angle DBC is equal to the angle BCD, the side BD is equal to the side DC; but h 6.1. BV was made equal to CA; therefore also CA is equal to CD, and the angle CDA equals to the angle DAC; therefore the angles CDA, DAC together, are double of the angle DAC: but BCD is equal to the angles CDA, DAC; therefore also BCD is double of DAC, and BCD is equal to each of the angles BDA, DBA ; each therefore of the angles BDA, DBA is double of the angle DAB; wherefore an isosceles triangle ABD is described, having each of the angles at the base double nf the third angle. Which was to be done.
PROP. XI. PROB.
To inscribe an equilateral and equiangular pent.:gon in a given circle.
Let ABCDE be the given circle; it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE.
Describe an isosceles triangle FGH, having cach of the a 10. 4. angles at G, H; double of the angle at F; and in the circle ABCDE inscribe b the triangle ACD equiangular to the tri-b2. 4. angle FGH, so that the angle CAD be equal to the angle at F, and each of the angles ACD, CDA equal to the an
В gle at G or H; wherefore each of the angles ACD, CDA is double of the angle
c9.1. CAD. Bisect the angles ACD, CDA by the straight lines CE, DB; and join AB, BC, CD, DE, EA: ABCDEG
H is the pentagon required:
Because each of the angles ACD, CDA is double of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another; but equal angles stand upon equal d circumferences; therefore the d 26. 3 five circamferences AB, BC, CD, DE, EA are equal to one
Book IV, another: and equal circumferences are subtended by equale
straight lines; therefore the five straight lines AB, BC, CD, © 29. 3, DE, EA are equal to one another. Wherefore the pentagon
ABCDE is equilateral. It is also equiangular; because the circumference AB is equal to the circumference DE: if to each be added BCD, the whole ABCD is equal to the whole EDCB: and the angle AED stands on the circumference
ABCD, and the angle BAE on the circumference EDCB; $ 27. 3. therefore the angle BAE is equal to the angle AED: for the
same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: therefore thc pentagon ABCDE is equiangular; and it has been shown that it is equilateral. Wherefore, in the given circle, an equilateral and equianguJar pentagon has been inscribed. Which was to be done.
PROP. XII. PROB.
To describe an equilateral and equiangular pentagon about a given circle.
a 11. 4.
Let ABCDE be the given circle; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE.
Let the angles of a pentagon, inscribed in a circle, by the last proposition, be in the points A, B, C, D, E, so that the circumferences AB,BC, CD, DE,EA are equala; and through the points A, B, C, D, E draw GH, HK, KL, LM, MG, touching b the circle ; take the centre F, and join FB, FK, FC, FL, FD : and because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre F, FC is perpendiculare to KL; therefore each of the angles at C is a right angle: for the same reason, the angles at the points B, D, are right angles: and because FCK is a right angle, the square of FK is equald to the squares of FC, CK: for the same reason, the square of FK is equal to the squares of FB, BK: therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the. square of FB; the remaining square of CK is therefore equal to
c 18. 3.
. 47. 1.
the remaining square of BK, and the straight line CK equal to Book IV. BK: and because FB is equal to FC, and FK common to the bar triangles BPK, CFK, the two BF, FK are equal to the two CF, FK; and the base BK is equal to the base KC; therefore the angle BFK is equal e to the angle KFC, and the angle BKF toe 8. d. FKC; wherefore the angle BFC is double of the angle KFC, and BKC double of FKC; for the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF: and be. cause the circumference BC is equal to the circumference CD, f 27.8. the angle BFC is equal to the
G angle CFD, and BFC is dou: ble of the angle KFC, and
E CFD double of CFL; therefore the angle KFC is equal to
M the angle. CFL ; and the right
F angle FCK: is equal to the right angle FCI: therefore, in the
two triangles EKC, FLC, there B are two anglès of one equal to two angles of the dther, each to each, and the side FC, which
KC L is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equals to the other sides, and the third angle to the third angle: g 26. L therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC; and because KC is equal to.CL, KL is double of KC: in the same manner, it may be shown that HK is double of BK: and because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL: in like manner, it may be shown * that GH, GM,ML are each of them equal to HK or KL: therefore the pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and KLM double of FLC, as was before demonstrated, the angle HKL is equal to KLM: and in like manner it may be shown, that each of the angles'KHG, HGM, GML is equal to the angle HKL or KLM: therefore the five angles GHK, HKL, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular: and it is equilateral, as was demonstrated; and it is described about the cirele ABCDE. Which was to be done.