PROPOSITION I. PROBLEM.

To bisect a given angle.

Let BAC be the given angle, it is required to bisect it.

In AB take any point D; from A, with the radius AD, describe an arc intersecting AC in E; join DE, bisect it in F (8. 1.), and join AF: the angle DAF is equal to the angle EAF.

Because the triangles DAF and EAF have the side DA equal to EA (Def. 54. 1.), DF equal to EF, by the construction, and FA common; the three sides of the one are equal to the three sides of the other, and therefore the triangles are equivalent (7. 1.); and the angle DAF is equal to the angle EAF; and therefore the straight line FA bisects the angle DAE, or BAC. Which was to be done.

PROP II. 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 (8. 1.) AB, AC in the points D, E, and from these points draw DF, EF at right angles (9. 1.) to AB, AC; 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: then, because AD is equal to DB, 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 shown, 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, shall pass through the extremities of the other two, and be described about the triangle ABC. Which was to be done.

COR. I. In the same way the circumference of a circle may be described through any three points, not in a straight line.

COR. II. If a segment of a circle is given, the circle may be described of which it is a segment.

PROP. III. PROB.

To inscribe a circle in a given triangle.

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

A

G

D

Bisect (1. 6.) the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, from which draw (10. 1.) DE, DF, DG perpendiculars to AB, BC, CA; and because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that 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 shall be equal (6. 2.); 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, shall pass through the extremities of the other two, and 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 (8. 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. Which was to be done.

B

PROP. IV. 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 (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 equila-
teral. It is also rectangular; for the B
straight line BD, being the diameter of
the circle ABCD, BAD is a semicircle;
wherefore the angle BAD is a right (22.
3.) angle; for the same reason each of
the angles ABC, BCD, CDA is a right
angle; therefore the quadrilateral figure

E

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.

PROP. V. PROB.

To describe a square about a given circle.

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

A

F

Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw (9. 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 (10. 3.) angles: for the same reason, the angles at the points B, C, D are right angles; and because the angle B AEB is a right angle, as likewise is EBG, GH is parallel (2. Cor. 2. 2.) 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 pa

E

D

H

K

rallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms, and GF is therefore equal (9. 2.) 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 (9. 2.) 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. Which was to be done.

PROP. VI. THEOR.

If from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half, and so on: there shall at length remain a magnitude less than the least of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on; there shall at length remain a magnitude less than C.

And

A

K

F

D

For C may be multiplied, so as at length to become greater than AB. Let it be so multiplied, and let DE its multiple be greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in AB be AK, KH, HB; and the divisions in ED be DF, FG, GE. because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C. Q. E. D. And if only the halves be taken away, the same thing may in the same way be demonstrated.

H

B

PROP. VII. THEOR.

Similar polygons inscribed in circles are to one another as the squares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles; as the square of BM is to the square of GN, so is the polygon ABCDE to the polygon FGHKL.

Join BE, AM, GL, FN: and because the polygon ABCDE is similar to the polygon FGHKL, and similar polygons are divided into similar triangles (20. 5.); the triangles ABE, FGL are similar and

equiangular (7. 5.); and therefore the angle AEB is equal to the an gle FLG: but AEB is equal (13. 3.) to AMB, because they stand upon the same circumference; and the angle FLG is, for the same reason, equal to the angle FNG: therefore also the angle AMB is equal to FNG: and the right angle BAM is equal to the right (22. 3.) angle GFN; wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another; and therefore as BM to GN so (5. 5.) is BA to GF, and therefore the duplicate ratio of BM to GN, is the same (16. 4.) with the duplicate ratio of BA to GF but the ratio of the square of BM to the square of GN, is the duplicate (20. 5.) ratio of that which BM has to GN; and the ratio of the polygon ABCDE to the polygon FGHKL, is the duplicate ratio of that which BA has to GF; therefore, as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore, similar polygons, &c. Q. E. D.

1

:

PROP. VIII. THEOR.

Circles are to one another as the squares of their diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters;

as the square of BD to the square of FH, so is the circle ABCD

to the circle EFGH.