AG, make the angle GAB equal to the angle DFE, and join BC. Therefore, because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal (32. 3.) to the angle ABC in the alternate segment of the circle : But HAC is equal to the angle DEF; therefore also the angle ABC is equal to DEF; for the same reason, the angle ACB is

equal to the angle DFE; therefore the remaining angle BAC is equal (4. Cor. 32. 1.) to the remaining angle EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC.

PROP. III. PROB.

About a given circle to describe a triangle equiangular to a given triangle.

Let ABC be the given circle and DEF the given triangle; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF.

Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draw any straight line KB; at the point K in the straight line KB, make (Prop. 23 1.) the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN, NCL touching (Prop. 17.3.) the circle ABC: Therefore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, the angles at the points A, B, C, are right (18. 3.) angles. And because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two triangles; and because two of

them, KAM, KBM, are right angles, the other two AKB, AMB are equal to two right angles: But the angles DEG, DEF are likewise equal (13.1.) to two right angles; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG; wherefore the remain

ing angle AMB is equal to the remaining angle DEF. In like manner, the angle LMN may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal (32. 1.) to the remaining angle EDF : Wherefore the triangle LMN is equiangular to the triangle DEF: and it is described about the circle ABC.

PROP. IV. PROB.

To inscribe a circle in a given triangle.

Let the given triangle be ABC; it is required to inscribe a circle in ABC.

E

A

G

D

Bisect (9. 1.) the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, from which draw (12. 1.) DE, DF, DG perpendiculars to AB, BC, CA. Then because the angle EBD is equal to the angle FBD, the angle ABC being bisected by BD; and because the right angle BED, is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal (26. 1.); wherefore DE is equal to DF. For the same 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 of them, will pass through the extremities of the other two, and will 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 (1 Cor. 16. 3.) the circle. Therefore the straight lines AB, BC, CA, do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC.

B

PROP. V. PROB.

To describe a circle about a given triangle.

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

Bisect (10. 1.) AB, AC in the points D, E, and from these points draw

DF, EF at right angles (11. 1.) to AB, AC; DF, EF produced will 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: then, because AD is equal to BD, and DF common, and at right angles to AB, the base AF is equal (4. 1.) to the base FB. In like manner,

it may be shewn that CF is equal to FA; and therefore BF 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 one of them, will pass through the extremities of the other two, and be described about the triangle ABC.

COR. 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; but when the centre is 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 falls without the triangle, the angle opposite to the side beyond which it is, 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 falls within it; if it be a right angle 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.

SCHOLIUM.

1. From the demonstration it is evident that the three perpendiculars bisecting the sides of a triangle, meet in the same point; that is, the centre of the circumscribed circle.

2. A circular segment arch of a given span and rise, may be drawn by a modification of the preceding problem.

Let AB be the span and SR the rise.

Join AR, BR, and at their respective points of bisection, M, N, erect

the perpendicular MO, NO to AR, BR; they will intersect at O, the centre of the circle. That OA=OR=OB, is proved as before.

The joints between the arch-stones, or voussoirs, are only continuations of radii drawn from the centre O of the circle.

R

MN

B

A

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 the diameters, AC, BD at right angles to one another, and join AB, BC, CD, DA; because BE is equal to ED, E being the centre, and

D

D

because EA is at right angles to BD, and common to the triangles ABE, ADE; the base BA is equal (4. 1.) to the base AD; 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 be- B ing a diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right angle (31. 1.); 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 shewn to be equilateral; therefore it is a square; and it is inscribed in the circle ABCD.

SCHOLIUM.

Since the triangle AED is right angled and isosceles, we have (Cor. 2. 47. 1) AD AE :: √2:1; hence the side of the inscribed square is to the radius, as the square root of 2, is to unity.

PROP. VII. PROB.

To describe a square about a given circle.

G

A

F

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 (17. 3.) FG, GH, HK, KF touching the circle; 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 (18. 3.); 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 (28. 1.) to AC; 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 (34. 1.) to HK, and GH to FK; and because AC is equal to BD, and also 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 (34. 1.) is likewise a right angle: in the same manner, it may be shewn 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.

B

H

E

D

K

PROP. 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.

E

Bisect (10. 1.) each of the sides AB, AD, in the points F, E, and through E draw (31. 1.) EH parallel to AB or DC, and through F draw FK parallel to AD or BC; therefore each of the figures, AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are equal (34. 1.); and because that 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; in the same manner 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, will pass through the extremities of the other three; and will also touch the straight lines AB, BC, CD, DA, because the angles at the points E, F, H, K, are right angles (29. 1.), and because the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle (16. 3.); therefore each of the straight lines AB, BC,

G

F

B

H

CD, DA touches the circle, which is therefore inscribed in the squares ABCD.

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; 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 (8. 1.) to the angle BAC, and the angle DAB is bisected by the straight line AC. In the same manner it may be demonstrated, 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 the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA: and the side EA (6. 1.) to the side EB. In the same manner, it may be demonstrated, that the straight

E

B

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, must pass