BOOK IV.

DEFINITIONS.

I. A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed.

2. Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed.

3. A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle.

4. A rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure. touches the circumference of the circle.

5. Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed.

6. A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed.

7. A straight line is said to be fitted into a circle when its extremities are on the circumference of the circle.

DEFINITIONS I-7.

I append, as usual, the Greek text of the definitions.

I. Σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφεσθαι λέγεται, όταν ἑκάστη τῶν τοῦ ἐγγραφομένου σχήματος γωνιῶν ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται.

2. Σχῆμα δὲ ὁμοίως περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἑκάστης γωνίας τοῦ, περὶ ὃ περιγράφεται, ἅπτηται.

3. Σχῆμα εὐθύγραμμον εἰς κύκλον ἐγγράφεσθαι λέγεται, ὅταν ἑκάστη γωνία τοῦ ἐγγραφομένου ἅπτηται τῆς τοῦ κύκλου περιφερείας.

4.

Σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφεσθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἐφάπτηται τῆς τοῦ κύκλου περιφερείας.

5. Κύκλος δὲ εἰς σχῆμα ὁμοίως ἐγγράφεσθαι λέγεται, ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται.

6. Κύκλος δὲ περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης γωνίας του, περὶ ὃ περιγράφεται, ἄπτηται.

7. Εὐθεῖα εἰς κύκλον ἐναρμόζεσθαι λέγεται, ὅταν τὰ πέρατα αὐτῆς ἐπὶ τῆς περιφερείας ᾖ τοῦ κύκλου.

In the first two definitions an English translation, if it is to be clear, must depart slightly from the exact words used in the Greek, where "each side" of one figure is said to pass through "each angle" of another, or "each angle". (i.e. angular point) of one lies on each side” of another (ἑκάστη πλευρά, ἑκάστη γωνία).

It is also necessary, in the five definitions 1, 2, 3, 5 and 6, to translate the same Greek word anτηraι in three different ways. It was observed on III. Def. 2 that the usual meaning of ἅπτεσθαι in Euclid is to meet, in contradistinction to ἐφάπτεσθαι, which means to touch. Exceptionally, as in Def. 5, άπτεσθαι has the meaning of touch. But two new meanings of the word appear, the first being to lie on, as in Deff. 1 and 3, the second to pass through, as in Deff. 2 and 6; “each angle" lies on (άπτεται) a side or on a circle, and “ each side,” or a circle, passes through (άπτεται) an angle or " each angle.” The first meaning of lying on is exemplified in the phrase of Pappus ἅψεται τὸ σημεῖον θέσει δεδομένης εὐθείας, “ will lie on a straight line given in position "; the meaning of passing through seems to be much rarer (I have not seen it in Archimedes or Pappus), but, as pointed out on III. Def. 2, Aristotle uses the compound ἐφάπτεσθαι in this sense.

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Simson proposed to read ἐφάπτηται in the case (Def. 5) where άπτηται means touches. He made the like suggestion as regards the Greek text of III. 11, 12, 13, 18, 19; in the first four of these cases there seems to be Ms. authority for the compound verb, and in the fifth Heiberg adopts Simson's correction.

BOOK IV. PROPOSITIONS

PROPOSITION I.

Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the

circle.

Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle;

thus it is required to fit into the circle ABC a straight line equal to the straight line D.

Let a diameter BC of the circle ABC be drawn.

Then, if BC is equal to D,

that which was enjoined will have

been done; for BC has been fitted into the circle ABC equal to the straight line D.

But, if BC is greater than D,

let CE be made equal to D, and with centre C and distance CE let the circle EAF be described;

let CA be joined.

Then, since the point C is the centre of the circle EAF,

CA is equal to CE.

But CE is equal to D;

therefore D is also equal to CA.

Therefore into the given circle ABC there has been fitted CA equal to the given straight line D.

Q. E. F.

Of this problem as it stands there are of course an infinite number of solutions; and, if a particular point be chosen as one extremity of the chord to be "fitted in," there are two solutions. More difficult cases of "fitting into" a circle a chord of given length are arrived at by adding some further condition, e.g. (1) that the chord is to be parallel to a given straight line, or (2) that the chord, produced if necessary, shall pass through a given point. The former problem is solved by Pappus (111. p. 132); instead of drawing the chord as a tangent to a circle concentric with the given circle and having as radius a straight line the square on which is equal to the difference between the squares on the radius of the given circle and on half the given length, he merely draws the diameter of the circle which is parallel to the given direction, measures from the centre along it in each direction a length equal to half the given length, and then draws, on one side of the diameter, perpendiculars to it through the two points so determined.

H

K

The second problem of drawing a chord of given length, being less than the diameter of the circle, and passing through a given point, is more important as having been one of the problems discussed by Apollonius in his work entitled vevoeis, now lost. Pappus states the problem thus (VII. p. 670): "A circle being given in position, to fit into it a straight line given in magnitude and verging (vevovσay) towards a given (point)." To do this we have only to place any chord HK in the given circle (with centre O) equal to the given length, take the middle point of it, with O as centre and OL as radius describe a circle, and lastly through the given point C draw a tangent to this circle meeting the given circle in A, B. AB is then one of two chords which can be drawn satisfying the given conditions, if C is outside the inner circle; if C is on the inner circle, there is one solution only; and, if C is within the inner circle, there is no solution. Thus, if C is within the outer (given) circle, besides the condition that the given length must not be greater than the diameter of the circle, there is another necessary condition of the possibility. of a solution, viz. that the given length must not be less than double of the straight line the square on which is equal to the difference between the squares (1) on the radius of the given circle and (2) on the distance between its centre and the given point.

A

B

PROPOSITION 2.

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

Let ABC be the given circle, and DEF the given triangle;

thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF.

Let GH be drawn touching the circle ABC at A [111. 16, Por.];

H. E. II.

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on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE;

[I. 23]

let BC be joined.

A

E

H

Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle,

therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle.

[III. 32] But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF. For the same reason

the angle ACB is also equal to the angle DFE ; therefore the remaining angle BAC is also equal to the remaining angle EDF.

[1.32]

Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle.

Q. E. F.

Here again, since any point on the circle may be taken as an angular point of the triangle, there are an infinite number of solutions. Even when a particular point has been chosen to form one angular point, the required triangle may be constructed in six ways. For any one of the three angles may be placed at the point; and, whichever is placed there, the positions of the two others relatively to it may be interchanged. The sides of the triangle will, in all the different solutions, be of the same length respectively; only their relative positions will be different.

This problem can of course be reduced (as it was by Borelli) to III. 34, namely the problem of cutting off from a given circle a segment containing an angle equal to a given angle. It can also be solved by the alternative method applicable to III. 34 of drawing "angles at the centre" equal to double the angles of the given triangle respectively; and by this method we can easily solve this problem, or 111. 34, with the further condition that one side of the