angle ADB is equal to the angle BCE, being in the same segment (Prop. XVIII. Cor. 1, Bk. III.); therefore the triangles ABD, BCE are similar; hence the proportion, AD: BD::CE: BC;

and, consequently,

ADX BC= BD X CE.

B

C

A

E

Again, since the angle ABE is equal to the angle CBD, and the angle BAE is equal to the angle B DC, being in the same segment (Prop. XVIII. Cor. 1, Bk. III.). the triangles A BE, B C D are similar; hence,

AB: AE:: BD: CD;

and consequently,

ABX CD = AEX BD.

By adding the corresponding terms of the two equations obtained, and observing that

BD×AE+BD× CE=BD (AE+CE)=BD× AC, we have

BD XAC ABX CD+ADX BC.

=

PROPOSITION XXXIX.-THEOREM.

292. The diagonal of a square is incommensurable with its side.

Let A B C D be any square, and A C its diagonal; then A C is incommensurable with the side A B.

To find a common measure, if there be one, we must apply AB, or its equal CB, to CA, as often as it can be done. In order to do this, from the point C as a center, with

Ꭰ

A G B

E

a radius C B, describe the semicircle F B E, and produce AC to E. It is evident that CB is contained once in AC,

with a remainder A F, which remainder must be compared with BC, or its equal, A B.

The angle ABC being a right angle, A B is a tangent to the circumference, and A E is a secant drawn from the same point, so that (Prop. XXXV.)

AF: AP::AB: AE.

Hence, in comparing A F with A B, the equal ratio of A B to A E may be substituted; but A B or its equal C F is contained twice in A E, with a remainder A F; which remainder must again be compared with A B.

Thus, the operation again consists in comparing A F with A B, and may be reduced in the same manner to the comparison of A B, or its equal CF, with AE; which will result, as before, in leaving a remainder A F; hence, it is evident that the process will never terminate; consequently the diagonal of a square is incommensurable with its side.

293. Scholium. The impossibility of finding numbers to express the exact ratio of the diagonal to the side of a square has now been proved; but, by means of the continued fraction which is equal to that ratio, an approximation may be made to it, sufficiently near for every practical purpose.

BOOK V.

PROBLEMS RELATING TO THE PRECEDING

BOOKS.

PROBLEM I.

294. To bisect a given straight line, or to divide it into

E

A

B

D

From the point A as a center, with a radius greater than the half of A B, describe an arc of a circle; and from the point B as a center, with the same radius, describe another arc, cutting the former in the points C and D. Through C and D draw the straight line CD; it will bisect A B in the point E.

For the two points C and D, being each equally distant from the extremities A and B, must both lie in the perpendicular raised from the middle point of AB (Prop. XV. Cor., Bk. I.). Therefore the line CD must divide the line A B into two equal parts at the point E.

PROBLEM II.

295. From a given point, without a straight line, to draw a perpendicular to that line.

Let A B be the straight line, and let C be a given point without the line.

From the point C as a center, and with a radius sufficiently great, describe an are cutting the line AB in two points, A and B; then, from the points A and B as centers, with a radius greater than half of A B, describe two arcs cutting each other in D, and draw the straight line CD; it will be the perpendicular required.

For, the two points C and D are each equally distant from the points A and B; hence, the line CD is a perpendicular passing through the middle of A B (Prop. XV. Cor., Bk. I.).

PROBLEM III.

296. At a given point in a straight line to erect a perpendicular to that line.

Let A B be the straight line, and let

D be a given point in it.

C

A

D

B

In the straight line AB, take the points A and B at equal distances from D; then from the points A and B as centers, with a radius greater than AD, describe two arcs cutting each other at C; through C and D draw the straight line CD; it will be the perpendicular required.

For the point C, being equally distant from A and B, must be in a line perpendicular to the middle of A B (Prop. XV. Cor., Bk. I.); hence CD has been drawn perpendicular to A B at the point D.

297. Scholium. The same construction serves for making a right angle, A D C, at a given point, D, on a given straight line, A B.

PROBLEM IV.

298. To erect a perpendicular at the end of a given straight line.

Let A B be the straight line, and B the end of it at which a perpendicular is to be erected.

D

C

B

From any point, D, taken without the line A B, with a radius equal to the distance D B, describe an arc cutting the line A B at the points A and B ; through the point A, and the center D, draw the diameter AC. Then through C, where the diameter meets the arc, draw the straight line CB, and it will be the perpendicular required. For the angle A B C, being inscribed in a semicircle, is a right angle (Prop. XVIII. Cor. 2, Bk. III.).

PROBLEM V.

299. At a point in a given straight line to make an angle equal to a given angle.

Let A be the given point, A B the given line, and EFG the given angle.

From the point F as

D

A A

GA

B

a center, with any radius, describe an arc, GE, terminating in the sides of the angle; from the point A as a centre, with the same radius, describe the indefinite arc BD. Draw the chord G E; then from B as a centre, with a radius equal to G E, describe an arc cutting the arc BD in C. Draw A C, and the angle CAB will be equal to the given angle EFG.

For the two arcs, BC and GE, have equal radii and equal chords; therefore they are equal (Prop. III. Bk.