therefore their other sides are equal; (1. 26) wherefore DE is equal to DF;

for the same reason, DG is equal to DF; therefore DE is equal to DG; (ax. 1)

wherefore the three straight lines DE, DF, DG, are equal to one another;

and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two.

And because the angles at the points E, F, G, are right angles,

and the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle; (III. 16)

therefore the straight lines AB, BC, CA, do each of them touch the circle;

and therefore the circle EFG is inscribed in the triangle Q. E. F.

ABC.

PROP. V.-PROBLEM.

To describe a circle about a given triangle.
(References-Prop. I. 4, 10, 11; III. 31.)

Let the given triangle be ABC.

It is required to describe a circle about ABC.

Bisect AB, AC, in the points D, E, (1. 10)

and from these points draw. DF, EF, at right angles to AB, AC. (1. 11.)

DF, EF, produced meet one another;

for, if they do not meet, they are parallel,

wherefore AB, AC, which are at right angles to them, are parallel; which is absurd;

let them meet in F, and join FA;

also if the point F be not in BC, join BF, CF;

and from the centre F, at the distance of BF, CF, or AF, describe the circle ABC.

Then the circle ABC shall be described about the triangle ABC.

DEMONSTRATION

Because AD is equal to DB, and DF common, and at right angles to AB,

therefore the base AF is equal to the base FB. (1. 4.)

In like manner, it may be shown that CF is equal to FA; therefore BF is equal to FC; (ax. 1)

wherefore FA, FB, FC, are equal to one another,

and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other two.

Therefore the circle ABC is described about the triangle
ABC.
Q. E. F.

Cor. And it is manifest that, when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle; (I. 31)

but when the centre is in one of the sides of the triangle, the angle opposite to this side is a right angle, being in a semicircle;

and if the centre fall without the triangle, the angle opposite to the side beyond which it is, is greater than a right angle, being in a segment less than a semicircle.

Wherefore, conversely, if the given triangle be acute-angled, the centre of the circle falls within it;

if it be a right-angled triangle, the centre is in the side opposite to the right angle;

and, if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

To inscribe a square in a given circle.

(References Prop. I. 4, 11; III. I. 31.)

Let ABCD be the given circle.

It is required to inscribe a square in ABCD.

B

A

CONSTRUCTION

Draw the diameters AC, BD, at right angles to one another; (III. 1,

and I. 11)

and join AB, BC, CD, DA.

Then the figure ABCD shall be inscribed in the circle ABCD.

DEMONSTRATION

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 AD; (1.4)

and for the same reason, BC, CD, are each of them equal to BA or AD;

therefore the quadrilateral figure ABCD is equilateral.

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; (III. 31)

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 the quadrilateral figure ABCD is a square; (1. def. 30) and it touches the circumferences of the circle at each of its angles. Therefore the square ABCD is inscribed in the circle ABCD. Q. E. F.

PROP. VII.- PROBLEM.

To describe a square about a given circle.
(References-Prop. 1. 28, 34; III. 17, 18.)

Let ABCD be the given circle.

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; (III. 17)

Then the figure GHKF, shall be described about the circle ABCD.

DEMONSTRATION

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; (III. 18)

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, therefore GH is parallel to AC; (L. 28)

for the same reason,

AC is parallel to FK;

and in like manner it may be demonstrated that GF, HK, are each of them parallel to BED;

therefore the figures GK, GC, AK, FB, BK, are parallelograms;

and therefore GF is equal to HK, and GH to FK; (1. 34)

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;

therefore GH, FK, are each of them equal to GF, or HK;

wherefore the quadrilateral figure GHKF is equilateral.

It is also rectangular;

for GBEA being a parallelogram, and AEB a right angle,

therefore AGB is likewise a right angle; (1. 34)

in the same manner, it may be shown that the angles at H, K, F, are right angles;

therefore the quadrilateral figure GHKF is rectangular ; and it was demonstrated to be equilateral;

therefore GHKF is a square;

and all its sides touch the circumference of the circle;

wherefore the square GHKF is described circle ABCD.

about the

PROP. VIII.-PROBLEM.

To inscribe a circle in a given square.

(References-Prop. I. 10, 29, 31, 34; III. 16, cor.)

Let ABCD be the given square.

It is required to inscribe a circle in ABCD.

Q. E. F.

CONSTRUCTION

Bisect each of the sides AB, AD, in the points F, E; (1. 10)

and through E draw EH parallel to AB or DC; (1. 31)

and through F draw FK parallel to AD or BC, cutting EH in G; from the centre G, at the distance GE, GF, or GK, describe the circle EFHK.

Then the circle EFHK shall be inscribed in the square ABCD.

DEMONSTRATION

Because each of the figures AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram (constr.)

their opposite sides are equal; (1. 34)

and because AD is equal to AB, (1. def. 30) and that AE is the half of AD, and AF the half of AB,

therefore AE is equal to AF; (ax. 7)

and the sides opposite to these are equal, wherefore FG is equal to GE;