or

AD or BC. therefore each of the figures AK, KB, AH, HD, AG, Book IV.
GC, BG, GDis a parallelogram, and their opposite sides are equal".
and because AD is equal to AB, and that AE is the half of AD,c. 34. s.
and AF the half of AB; AE is equal to

А E
AF. wherefore the sides opposite to these

D
are equal, viz. FG to GE. in the same
manner it may be demonstrated that GH,

G
GK are each of them equal to FG or

K
GE. therefore the four straight lines
GE, GF, GH, GK are equal to one ano-
ther; and the circle described from the
center G, at the distance of one of them

B H H C
shall pass thro’the extremities of the other
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 d. 29. 6.
straight line which is drawn from the extremity of a diameter, at
right angles to it, touches the circle e, therefore each of the e. 16. 5.
straight lines AB, BC, CD, DA touches the circle, which there-
fore is inscribed in the square ABCD. Which was to be done.

To describe a circle about a given square.

.

a. 8. I,

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 two BA, AC;
and the base DC is equal to the base BC;
wherefore the angle DAC is equal a to the
angle BAC, and the angle DAB is bisected

E
by the straight line AC. in the same man-
ner it may be demonstrated that the angles
ABC, BCD, CDA are severally bisected by

B
the 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 fide EA is equal 5 to the b. 6.se
side EB. in the same manner it may be demonstrated that the straight

Book iv. 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 center E, at the distance of one of them, shall pass thro' the extremities of the other three, and bę described about the square ABCD. Which was to be done,

PROP. X. PROB.

To describe an ifosceles triangle

, having

each of the angles at the base double of the third angle.

Take any straight line AB, and divide • it in the point C, fo that the rectangle AB, BC be equal to the square of CA; and from the center A, at the distance AB describe the circle BDE, 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 about the triangle ADC describe 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 from
the point B without the circle

E
ACD two straight lines BCA, BD
are drawn to the circumference,
one of which cuts, and the other
meets the circle, and that the rec-
tangle AB, BC contained by the

A
whole of the cutting line, and the
part of it without the circle, is e-
qual to the square of BD which
meets it ; the straight line BD
touches the circle ACD. and
because BD touches the circle, BD
and DC is drawn from the point
of contact D, the angle BDC is equal 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 angles
CDA, DAC. but the exterior angle BCD is equal to the angles
CDA, DAC; therefore allo BDA is equal to BCD. but BDA is

d. 37. 3.

e. 32. 3.

f. 32. .

equal 6 to the angle CBD, because the side AD is equal to the Book IV. fide AB ; therefore CBD, or DBA is equal to BCD; and consequently the three angles BDA, DBA, BCD are equal to one ano- 8. 5. 1. ther. and because the angle DBC is equal to the angle BCD, the fide BD is equal b to the side DC. but BD was made equal to CA, h. 6. 1. therefore alfo CA is equal to CD, and the angle CDA equal 6 to 8. 5. 1. the angle DAC. therefore the angles CDA, DAC together, are double of the angle DAC. but BCD is equal to the angies 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.

PRO P. XI. PROB.
"O inscribe an equilateral and cquiangular penta-

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 each of the angles a. 10. 4. at G, H double of the angle at F; and in the circle ABCDE infcribe'o the triangle ACD equiangular to the triangle FGH, fo b. 2. 4. that the angle CAD be equal

A to the angle at F, and each of the angles ACD, CDA equal

F to the angle at Gor H; where

B

E fore each of the angles ACD, CDA is double of the angle CAD. Bisect - the angles ACD, CDA by the straight lines CE, DB, and join AB,

H BC, DE, EA. ABCDE is the pentagon required.

Because each of the angles ACD, CDA is double of CAD, and are bifected 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 circumferences; therefore the five circumferen- d. 26. 3. ces AB, BC, CD, DE, EA are equal to one another. and equal

C. 9. 1.

Book IV. circumferences are subtended by equal o straight lines; therefore

the five straight lines AB, BC, CD, DE, EA are equal to one
another. Wherefore the pentagon ABCDE is equilateral. It is
also equiangular ; because the

A
circumference AB is equal to
the circumference DE, if to

F
'each be added BCD, the whole

В.
ABCD is equal to the whole
EDCB. and the angle AED
stands on the circumference
ABCD, and the angle BAE
therefore the angle BAE is e-

G H
F. 27. 3. qualf 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 sewn that it is equilateral. Wherefore in the given circle an equilateral and equiangular pentagon has been inscribed. Which was to be done."

PRO P. XII. PROB.
O describe an equilateral and equiangular penta-

gon about a given circle.

2. 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 the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumfe

rences AB, BC, CD, DE, EA are equal"; and thro' the points A, Þ. 17. 3. B, C, D, E draw GH, HK, KL, LM, MG touching the circle;

take the center F, and join FB, FK, FC, FL, FD. and because the

straight line KL touches the circle ABCDE in the point C, to c. 18. 3.

which FC is drawn from the center F, FC is perpendicular to KL; therefore each of the angles at C is a right angle. for the same rea

son, the angles at the points B, D are right angles. and because FCK d. 47, I.

is a right angle, the square of FK is equal 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 CKis therefore equal to the remaining square

f. 27. 3.

of BK, and the straight line CK equal to BK. and because FB is Book IV.
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 to the anglec. 8.1.
KFC, and the angle BKF to FKC. wherefore the angle BFC is
double of the angle KFC, and BKC double of FKC. for the
fame 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

G
is equal to the angle CFD. and
BFC is double of the angle KFC,

А

E
and CFD double of CFL; there-
fore the angle KFC is equal to the H

M
angle CFL ; and the right angle

F
FCK is equal to the right angle
FCL. therefore in the two triangles B
FKC, FLC, there are two angles
of one equal to two angles of the

K
other, each to each, and the side
FC, which is adjacent to the equal angles in each, is common to
both; therefore the other sides shall be equal & to the other sides, 5. 26. 1.
and the third angle to the third angle, therefore the straight line
KC is equal to CL, and the angle FKC to the angle FLC. and be-
caufe KC is equal to CL, KL is double of KC. in the same man-

may be shewn 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 shewn 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 FRC, and KLM double of FLC, as was before demonstrated; the angle HKL is equal to KLM. and in like manner it may be shewn, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM. therefore the five angles GHK, HKI, 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.

ner, it

!