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Book IV. demonstrated, that the straight linés 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.

a 11. 2. 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, b 1. 4. in which place b the straight line BD equal to AC, which is not

greater than the diameter of the circle BDE; join DA, DC, and c 5.4. about the triangle ADC describe c 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
from the point B, without the
circle ACD, two straight lines
BCA, BD are drawn to the cir-
cumference, one of which cuts,
and the other meets the circle,
and that the rectangle AB, BC

A
contained by the whole of the
cutting line, and the part of it
without the circle, is equal to the

square of BD which meets it; d 37.3. the straight line BD touches d

the circle ACD; and because
BD touches the circle, and DC

D
is drawn from the point of con-

B e 32. 3. tact D, the angle BDC is equal e to the angle DAC in the

alternate segment of the circle; to each of these add the angle

CDA; therefore the whole angle BDA is equal to the two f 32. 1. angles CDA, DAC; but the exterior angle BCD is equalf to

the angles CDA, DAC; therefore also BDA is equal to-BCD;

but BDA is equal 8 to the angle CBD, because the side AD Book IV. is equal to the side AB; therefore CBD, or DBA is equal to BCD; and consequently the three angles BDA, DBA, BCD g 5. 1. are equal to one another; and because the angle DBC-is equal to the angle BCD, the side BD is equal b to the side DC; but h 6.1 BD was made equal to CA; therefore also CA is equal to CD, and the angle CDA equal 8 to the angle DAC; therefore the angles CDĂ, 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 of the third angle. Which was to be done.

PROP. XI. PROB.

TO inscribe an equilateral and equiangular pentagon 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 a an isosceles triangle FGH, having each 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- b 2. 4. angle FGH, so that the angle

А
CAD be equal to the angle
at F, and each of the angles
ACD, CDA equal to the

F

E

B angle at G or H ; wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect c the angles

c 9.1. ACD, CDA by the straight lines CE, DB; and join AB, BC, DE, EA. ABCDE is

G

H 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 circumferences AB, BC, CD, DE, EA are equal to one

Book IV. another : and equal circumferences are subtended by equal e

s straight lines ; therefore the five straight lines AB, BC, CD, e 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; therefore the 27. 3. angle BAE is equal f to the angle AED: for the same reason,

each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: therefore the pentagon ABCDE is equiangular; and it has been shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done.

PROP. XII. PROB.

TO describe an equilateral and equiangular pentagon about a given circle.

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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 the circle, by the

last proposition, be in the points_A, B, C, D, E, so that the a 11. 4. circumferences AB, BC, CD, DE, EA are equal a ; and through

the points A, B, C, D, E draw GH, HK, KL, LM, MG, b 17.3. touching 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 c 18. 3. centre F, FC is perpendicular c 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 d 47. 1. right angle, the square of FK is equal d 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 te

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 triangles BFK, 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 to e 8. 1.. 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 because the circumference BC is equal to the circumference CD, the angle BFC is equalf to the

f 27. 3. angle CFD; and BFC is double of the angle KFC, and CFD double of CÉL; therefore the

А

E angle KFC is equal to the angle CFL; and the right angle FCK H н

M M is equal to the right angle FCL:

F therefore, in the two triangles FKC, FLC, there are two angles of one equal to two angles

D

B of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is

K с L common to both; therefore the other sides shall be equal 8 to the other sides, and the third angle 8 26. 1. to the third angle: 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 circle ABCDE. Which was to be done.

P

Book IV.

PROP. XIII. PROB.

TO inscribe a circle in a given equilateral and equi. angular pentagon.

Let ABCDE be the given equilateral and equiangular penta

gon; it is required to inscribe a circle in the pentagon ABCDE. a 9.1.

Bisecta the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE: therefore, since BC is equal to CD, and CF common to the triangles BCF, DCF, the two sides BC, CF are equal

to the two DC, CF; and the angle BCF is equal to the angle b 4. 1. DCF; therefore the base BF is equal b to the base FD, and the

ather angles to the other angles, to which the equal sides are op-
posite; therefore the angle CBF is equal to the angle CDF : and
because the angle CDE is double of CDF, and that CDE is equal
to CBA, and CDF to CBF; CBA
is also double of the angle CBF;

A
therefore the angle ABF is equal
to the angle CBF; wherefore the

G

M M
angle ABC is bisected by the
straight line BF: in the same B
manner, it may be demonstrated,
that the angles BAE, AED are

bisected by the straight lines AF,
c 12. 1. FE: from the point F draw c H Η

FG, FH, FK, FL, FM perpen-
diculars to the straight lines AB,
BC, CD, DE, EA: and be-

C KD
cause the angle HCF is equal to
KCF, and the right angle FHC equal to the right angle FKC;
in the triangles FHC, FKC there are two angles of one equal
to two angles of the other, and the side FC, which is opposite

to one of the equal angles in each, is common to both; therefore d 26. 1. the other sides shall be equal d, each to each; wherefore the

perpendicular FH is equal to the perpendicular FK: in the same manner it may be demonstrated that FL, FM, FG are each of them equal to FH or FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other four, and

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