centre falls without the triangle, beyond the side opposite

to the obtuse angle.

PROPOSITION VI.

PROB.—To inscribe a square in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in ABCD.

Draw the diameters AC, BD at right angles to one another, and join AB, BC, CD, DA.

Because BE is equal to ED, for

E is the centre, and that EA is
common, and at right angles to BD;
the base BA is equal to the base B
AD:

*

And, for the same reason, BC, CD are each of them equal to BA, or AD; therefore the quadrilateral figure ABCD is equilateral.

D

4.1.

E

It is also rectangular; for the straight line BD being the diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right† angle:

For the same reason, each of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular: and it has been shown to be equilateral; therefore it is a square; and it is inscribed in the circle ABCD. Which was to be done.

PROPOSITION VII.

PROB.-To describe a square about a given circle.

Let ABCD be the given circle; it is required to describe a square about it.

Draw two diameters AC, BD of the circle ABCD, at right angles to one another; and through the points A, B, C, D, draw* FG, GH, HK, KF touching the circle:

B

† 31. 3.

G A

F

* 17. 3.

E

And because FG touches the circle
ABCD, and EA is drawn from the centre
E to the point of contact A, the angles at
A are right angles; for the same reason
the angles at the points B, C, D are right angles :
And because the angle AEB is a right angle, as likewise

is EBG; GH is parallel to AC:

H

C K

+ 18. 3.

I 28. 1.

34. 1.

§ 34. 1.

10. 1. + 31. 1.

34. 1.

|| 29. 1.

$16.3.

For the same reason AC is parallel to FK: and in like manner GF, HK may each of them be demonstrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal || to HK, and GH to FK:

And because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK; GH, FK are each of them equal to GF, or HK; therefore the quadrilateral figure FGHK is equilateral.

It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB § is likewise a right angle:

In the same manner, it may be shown that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular: and it was demonstrated to be equilateral; therefore it is a square; and it is described about the circle ABCD. Which was to be done.

PROPOSITION VIII.

PROB. To inscribe a circle in a given square.

Let ABCD be the given square; it is required to inscribe a circle in ABCD.

Bisect each of the sides AB, AD in the points F, E, and through E draw† EH parallel to AB or DC, and through F draw FK parallel to AD or BC;

F

A E

D

G

K

Therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram; and their opposite sides are equal: and because AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite to these are equal, viz. FG to GE:

B

H

In the same manner it may be demonstrated that GH, GK are each of them equal to FG or GE; therefore 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 three, and touch the straight lines AB, BC, CD, DA; because the angles at the points, E, F, H, K, are right|| angles, and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle §; therefore each of the straight lines AB, BC, CD, DA touches the circle, which therefore is inscribed in the square ABCD. Which was to be done.

PROPOSITION IX.

PROB. 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:

E

D

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, each A to each; and the base DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bisected by the straight line AC:

*

B

In the same manner, it may be demonstrated that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC;

* 8. 1.

Therefore, because 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 EA is equal to the side + 6.1. EB:

In the same manner, it may be 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.

PROPOSITION X.

PROB. To describe an isosceles 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 * 11. 2. 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 † 1. 4. to AC, which is not greater than the diameter of the circle BDE; join DA, DC, and about the triangle ADC describe‡ ‡ 5. 4. 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.

| 37. 3.

§ 32.3.

82. 1.

5.1.

↑ 6. 1.

† 5. 1.

32. 1.

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 ACD, two straight

lines BCA, BD are drawn to the
circumference, one of which cuts,
and the other meets the circle, and
that the rectangle AB, BC, con-
tained by the whole of the cutting
line, and the part of it without the

E

A

B

circle, is equal to the square of BD which meets it, the straight line BD touches || the circle ACD:

And because BD touches the circle, 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 also BDA is equal to BCD:

But BDA is equal* to the angle CBD, because the side AD is equal to the side AB; therefore CBD, or DBA, is equal to BCD; and consequently the three angles BDA, DBA, BCD are equal to one another;

And because the angle DBC is equal to the angle BCD, the side BD is equal to the side DC: but BD was made equal to CA, therefore also CA is equal to CD, and the angle CDA equal‡ 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 of the third angle. Which was to be done.

PROPOSITION 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* an isosceles triangle FGH, having each of the • 10. 4.

angles at G, H, double of the angle at F; and in the circle ABCDE, inscribet the triangle ACD, equiangular to the triangle FGH, so that the angle CAD be equal to the

angle at F, and each of G

H

the angles ACD, CDA equal to the angle at G

D

or H; wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect the angles ACD, CDA by the ‡ 9. 1. straight lines CE, DB; and join AB, BC, DE, EA: ABCDE 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 || circumferences; || 26.3. therefore the five circumferences, AB, BC, CD, DE, EA are equal to one another:

And equal circumferences are subtended by equal§ § 29. 3. 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 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 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 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.

PROPOSITION XII.

PROB. To describe an equilateral and equiangular pentagon about a given circle.

Let ABCDE be the given circle; it is required to

K

¶ 27. 3.