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 because the angles at the points E, F, H, K, are right angles; (1. 29)

and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle; (II. 16, Cor.) therefore each of the straight lines AB, BC, CD, DA, touches the circle EFHK,

wherefore the circle EFHK is inscribed in the square ABCD. Q. E. F.

PROP. IX.-PROBLEM.

To describe a circle about a given square.

(References-Prop. I. 6, 8.)

Let ABCD be the given square.

It is required to describe a circle about ABCD.

B

D

CONSTRUCTION

Join AC, BD, cutting one another in E;

then, from the centre E, at the distance of EA, EB, EC, or ED, describe the circle ABCD.

Then the circle ABCD shall be described about the square ABCD.

DEMONSTRATION

Because, in the two triangles, DAC, BAC, DA is equal to AB (1. def. 30), and AC common,

the two sides DA, AC, are equal to the two BA, AC, each to each; and the base DC is equal to the base BC;

therefore the angle DAC is equal to the angle BAC; (1.8)

and the angle DAB is bisected by the straight line AC.

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

Hence, because the angle DAB is equal to the angle ABC, (1. def. 30) 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; (ax. 7)

therefore the side EA is equal to the side EB. (1. 6.)

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 the extremities of the other three ;

wherefore the circle ABCD is described about the square ABCD. Q. E. F.

PROP. X.- PROBLEM.

To describe an isosceles triangle, having each of the angles at the base double of the third angle.

(References-Prop. 1. 5, 6, 32; II. 11; III. 32, 37; Iv. 1, 5.)

CONSTRUCTION

Take any straight line AB, and divide it in the point C, so that the rectangle AB, BC, may be equal to the square of CA; (11. 11)

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; (IV. 1) and join DA.

Then the triangle ABD shall be such as is required, that is, each of the angles ABD, ADB shall be double of the angle BAD.

E

Join DC, and about the triangle ADC describe the circle ACD. (IV. 5.)

DEMONSTRATION

Because the rectangle AB, BC, is equal to the square of AC, and that AC is equal to BD, (constr.)

the rectangle AB, BC, is equal to the square of BD; (ax. 1)

and because from the point B without the circle ACD, two straight lines BCA, BD, are drawn to the circumferences, one of which cuts, and the other meets the circle,

and that the rectangle AB, BC, 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;

the straight line BD touches the circle ACD; (III. 37)

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; (III. 32)

to each of these add the angle CDA;

therefore the whole angle BDA is equal to the two angles CDA, DAC; (ax. 2)

but the exterior angle BCD is equal to the angles CDA, DAC; (1.32)

therefore also BDA is equal to BCD; (ax. 1)

but BDA is equal to the angle CBD, (1. 5) because the side AD is equal to the side AB;

therefore CBD, or DBA, is equal to BCD; (ax. 1)

and consequently the three angles BDA, DBA, BCD, are equal to one, another.

Again, because the angle DBC is equal to the angle BCD,

the side DB is equal to the side DC; (1. 6)

but BD was made equal to CA;

therefore also CA is equal to CD; (ax. 1)

and the angle CDA is equal to the angle DAC; (1. 5)

therefore the angles CDA, DAC, together are double of the angle DAC;

but BCD is equal to the angles CDA, DAC; (1. 32)

therefore also BCD is double of DAC;

and BCD was proved to be equal to each of the angles BDA, DBA,

therefore each of the angles BDA, DBA, is double of the angle DAB.

Wherefore an isosceles triangle ABD has been described, having each of angles at the base double of the third angle. Q. E. F.

PROP. XI.-PROBLEM.

To inscribe an equilateral and equiangular pentagon in a given circle. (References-Prop. 1. 9; III. 26, 27, 29; Iv. 2, 10.)

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 at G, H, double of the angle at F; (IV. 10)

and in the circle ABCDE inscribe the triangle ACD, equiangular to the triangle FGH, (1v. 2)

so that the angle CAD may be equal to the angle at F, and each of the angles ACD, CDA, equal to the angle at G or H;

wherefore each of the angles ACD, CDA, is double of the angle CAD; bisect the angles ACD, CDA, by the straight lines CE, DB; (1.9) and join AB, BC, DE, EA.

Then ABCDE shall be the equilateral and equiangular pentagon required.

DEMONSTRATION

Because each of the angles ACD, CDA, is double of CAD, and that they 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; (IIL 26)

therefore the five circumferences AB, BC, CD, DE, EA, are equal to one another

and equal circumferences are subtended by equal straight lines; (III. 29)

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;

for, 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; (ax. 2)

but 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; (ш. 27)

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.

Q. E. F.

PROP. XII.-PROBLEM.

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

(References-Prop. 1. 4, 8, 26, 47; III. 17, 18, 27; Iv. 11.)

Let ABCDE be the given circle.

It is required to describe an equilateral and equiangular pentagon about the circle ABCDE.

Let the angular points of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumferences AB, BC, CD, DE, EA, are equal; (IV. 11)