right angle DEB is equal to the right angle DFB, the two triangles DEB, DFB have two angles of the one equal to two angles of the other, each to each, and the side DB, which is opposite to one of the equal angles in each, common to both-therefore their other sides are equal, and therefore DE is equal to DF: For the like reason, DG is equal to DF: Therefore 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 one of them, shall pass through the extremities of the other two, and shall touch the straight lines AB, BC, CA, 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 (3. 16): Wherefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the given triangle ABC.

Q. E. F.

PROP. V. PROB.

To describe a circle about a given triangle.

Let ABC be the given triangle: it is required to describe a circle about the triangle ABC.

Bisect AB, AC, in the points D, E, and from D, E, draw DF, EF at right angles to AB, AC; DF, EF

produced, will meet one another; (for, if they do not meet, they are parallel, and therefore AB, AC, which are at right angles to them, are parallel-which is

N

absurd ;) let them meet in F, and join FA; also, if the point F be not in BC, join BF, CF.

Then, because AD is equal to BD, and DF common, and at right angles to AB, therefore the base FA is equal to the base FB: In like manner, it may be shewn that FA is equal to FC; and therefore FB is equal to FC, and FA, FB, FC are equal to one another: Wherefore the circle described from the centre F, at the distance of any one of them, shall pass through the extremities of the other two, and be described about the given triangle ABC.

Q. E. F.

COR. And it is manifest, that if the centre of the circle fall within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle; but, if the centre be in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and if the centre fall without the triangle, the angle opposite to the side beyond which it falls, being in a segment less than a semicircle, is greater than a right angle: Wherefore, if the given triangle be acute-angled, the centre of the circle will fall within it; if it be right-angled, the centre will be in the side opposite to the right angle; and if it be obtuse-angled, the centre will fall without the triangle, beyond the side opposite to the obtuse angle.

PROP. 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 two diameters AC, BD, intersecting one another at right angles in E, and join AB, BC, CD, DA: Then,

because BE is equal to ED (for E is the centre), and that EA is common, and at right angles to BD-therefore the base AB is equal to the base AD: And, for the like reason, BC, CD are each of them equal to BA, or AD: B And 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, and therefore the angle BAD is a right angle (3. 31): For the like reason, each of the angles ABC, BCD, CDA is a right angle: And therefore the quadrilateral figure ABCD is rectangular: And it has been shewn to be equilateral; therefore it is a square, and it is inscribed in the given circle ABCD.

Q. E. F.

PROP. 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.

G A F

E

H C

K

D

Draw two diameters AC, BD, at right angles to one another, and through the points A, B, C, D, draw FG, GH, HK, KF touching the circle (3. 17): Then, because FG touches the B circle ABCD, and EA is drawn from the centre E to the point of contact A, the angles at A are right angles (3.18): For the like reason, the angles at the points B, C, D are right angles: And because the angle AEB is a right angle, as also the angle EBG, therefore GH is parallel to AC (1. 28): For the like reason, AC is parallel to FK: And, in like manner, GF, HK may be shewn to be, each of them, parallel to BED: Therefore the figures GK, AH, AK, BF, 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, therefore GH, FK are each of them equal to GF, or HK; and therefore the quadrilateral figure FGHK is equilateral: It is also rectangular; for, AEBG being a parallelogram, and AEB a right angle, therefore also AGB is a right angle (1. 34) And, in like manner, it may be shewn that the angles at F, H, K are right angles; therefore the quadrilateral figure FGHK is rectangular: And it was shewn to be equilateral; therefore it is a square, and it is described about the given circle ABCD.

PROP. VIII. PROB.

To inscribe a circle in a given square.

Q.E. F.

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

D

BH

Then 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, therefore AE is equal to AF, and therefore also the sides opposite to these are equal, viz. FG to GE In like manner it may be shewn that GH, GK are each of them equal to GF 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 any 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 (3. 16): Therefore each of the straight lines AB, BC, CD, DA touches the circle, which is therefore inscribed in the square ABCD.

Q.E. F.

PROP. 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: Then, because AB is equal to AD, and AC common to the two triangles BAC, DAC, the two sides BA,

B

E

AC are equal to the two DA, AC, each to each, and the base BC is equal to the base DC-therefore the angle BAC is equal to the angle DAC, and the angle BAD is bisected by the straight line AC: In like manner, it may be shewn that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC: 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, therefore the angle EAB is equal to the angle EBA, and the side EA to the side EB: And, in like manner, it may be shewn 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 any one of them, shall pass through the extremities of the other three, and be described about the given square ABCD. Q.E. F.