EUCLID'S ELEMENTS.

BOOK IV.

DEFINITIONS.

1. A rectilineal figure is said to be inscribed in a rectilineal figure, when every angle of the inscribed figure touches every side of the figure in which it is inscribed.*

2. In like manner a figure is said to be circumscribed about a figure, when all the sides of the circumscribed figure touch all the angles of that figure about which it is circumscribed.

5. In like manner a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle. touches every side of the figure in which it is inscribed.

*When a figure is within another, so that all the angles of the inner figure are upon the sides of the figure in which it is, this figure is said by the Greeks, yypάpɛoda, to be inscribed within the other, and the outward figure is said, Teрryρúpeσlar, to be circumscribed about the inner one; but when it is merely to describe a circle, as in the 25th prop. lib. 3, πpoσavαɣρά is used. This distinction is also observed in Ptolemy's Meyan Zurtagis, as may be seen in the ninth chapter of the First Book.

6. A circle is said to be circumscribed about a figure, when the circumference of the circle touches every angle of the figure about which it is circumscribed.

7. A right line is said to be applied in a circle when its extremities are in the circumference of the circle.

PROPOSITION I.

PROBLEM.

In a given circle to apply a right line equal to a given right line not greater than the diameter of the circle."

A

Let ABC be the given circle, and D the given right line not greater than the diameter of the circle; it is required in the circle ABC to apply a right line equal to the right line D. Draw BC the diameter of the circle ABC. If, therefore, BC is equal to D, what was proposed will now be done. For in the circle ABC, BC is applied equal to the right line D. But if BC is greater than D, make CE equal to D, and with centre c, and distance CE, describe the circle AEF and join CA. Therefore, because the point c is the centre of the circle AEF, CA is equal to

F

D

E

a 3. 1.

CE. But CE is equal to D; therefore D also is equal to CA. Therefore in the given circle ABC, &c. Q. E. F.

PROPOSITION II.

PROBLEM.

In a given circle to inscribe a triangle equiangular to a given triangle.

A

Н

Let ABC be the given circle, and DEF the given triangle; it is required in the circle ABC to inscribe a triangle equiangular to the triangle DEF. Draw GH touching the circle ABC in the point A, and at the right line GH, and at the point A in it, make the angle HAC equal to the angle DEF; again, at the right line GA, and at the point a in it, make the angle GAB equal E to the angle FDE, and join BC.

F

b 15 Def. 1.

a 23. 1.

Therefore, because some right line HG touches the circle ABC, and from the point of contact at A, ac is drawn in the circle, whence the angle HAC is equal to the angle ABC in the alternate segment of the circle. ↳ 32. 3. But the angle HAC is equal to DEF. For the same reason, the angle ACB is equal to the angle FDE, and

€ 32. 1.

therefore the remaining angle BAC is equal to the remaining angle EFD. Therefore the triangle ABC is equiangular to the triangle DEF, and is inscribed in 3 Def. 4. the circle ABC." Wherefore in a given circle, &c. Q. E. F.

a 1. 3.

PROPOSITION III.

PROBLEM.

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

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

Produce EF both ways to the points G, H, and find K the centre of the circle ABC, also draw anyhow the right line KB; and at the right line KB, and at the point K in it, make the angle BKA equal to the

b 23. 1.

angle DEG, also the an

c 17 3.

18.3.

e 32. 1.

f 13. 1.

gle BKC equal to DFH,
and through the points A,

M

L

K

D

B, C, draw the right lines LAM, MBN, NCL, touching the circle ABC.C

d

And because the lines LM, MN, NL, touch the circle ABC in the points A, B, C, and KA, KB, KC, are joined; the angles at the points A, B, C, are right angles. And because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it may be divided into two triangles, of which the angles MAK, KBM, are right ones; therefore the remaining angles AMB, AKB, are equal to two right ones; but DEG, DEF, are also equal to two right angles; therefore AKB, AMB, are equal to DEG, DEF, of which AKB is equal to DEG; whence the remaining angle AMB is equal to the remaining angle DEF. In like manner, it may be shown that LNM is equal to DEF; and therefore the remaining angle MLN is also equal to the remaining angle EDF. Therefore LMN is a triangle equiangular to the triangle DEF. Therefore about a given circle, &c. Q. E. F.

PROPOSITION IV.

PROBLEM.

To inscribe a circle in a given triangle.

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

a

Bisect the angles ABC, ACB, by the right lines BD, CD, and let them meet one another in the point D, 9. 1. and draw from the point D to AB, BC, CA, the perpendicular right lines DE, DF, DG.

b 12. 1.

C

And because the angle ABD is equal to the angle DBC, and BED is a right angle, consequently equal to the right angle BFD; therefore EBD, FBD, are two triangles, having two angles equal to two angles, and one side to one side; viz. the side BD opposite to one of the equal angles common to them both, and, therefore, the remaining sides of the one shall be equal to the remaining sides of the other; whence DE is equal 26. 1. to DF. For the same reason, DG is equal to DF. Therefore the three right lines DE, DF, DG, are equal to one another; wherefore from the centre D, and with the distance any one of them D DE, DF, DG, the circle described will pass through the remaining points, B and will touch the right lines AB, BC,

d

E

A

F

d

CA, wherefore the angles at the points E, F, G, are right ones. For if it cut them, a line drawn at right angles to the diameter of the circle from the extremity will fall within the circle, which has been shown to be absurd. Therefore, with the centre D, and distance any 16. S. one of them DE, DF, DG, the circle described will not cut the right lines AB, BC, CA; whence it touches them, and the circle will be inscribed in the triangle Therefore in a given triangle, &c. Q. E. F.

ABC.e

Deductions.

1. The three right lines which bisect the three angles of a triangle meet in the same point.

2. If a circle be inscribed in a right angled triangle, the excess of the two sides containing the right angle above the third side is equal to the diameter of the inscribed circle.

e 5 Def. 4.