Let AB, CD be two chords, which intersect each other at E; then will

AE: DE::EC: EB.

Join AC and BD. In the triangles AEC, BED, the angles at E are equal, being vertical angles (Prop. IV. Bk. I.); the angle A is equal to

A

E

the angle D, being measured by half the same arc, BC (Prop. XVIII. Cor. 1, Bk. III.); for the same reason, the angle C is equal to the angle B; the triangles are therefore similar (Prop. XXII.), and their homologous sides give the proportion,

AE: DE::EC: EB.

=

283. Cor. Hence, A EX EB DEXEC; therefore the rectangle of the two segments of the one chord is equal to the rectangle of the two segments of the other.

PROPOSITION XXXIV. - THEOREM.

284. If from the same point without a circle two secants be drawn, terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments.

Let EB, E C be two secants drawn from the point E without a circle, and terminating in the concave arc at the points B and C; then will

EB:EC::ED: EA.

For, joining A C, BD, the triangles AEC, BED have the angle E common; and the angles B and C, being

B

A

E

measured by half the same arc, A D, are equal (Prop. XVIII. Cor. 1, Bk. III.); these triangles are therefore similar (Prop. XXII. Cor.), and their homologous sides give the proportion,

EB: EC::ED: EA.

EA =

285. Cor. Hence, EB X E A ECXED; therefore the rectangle contained by the whole of one secant and its external segment is equivalent to the rectangle contained by the whole of the other secant and its external segment.

PROPOSITION XXXV.-THEOREM.

286. If from a point without a circle there be drawn a tangent terminating in the circumference, and a secant terminating in the concave arc, the tangent will be a mean proportional between the whole secant and its external segment.

From the point E let the tangent EA, and the secant E C, be drawn; then will E C: EA:: EA: ED.

For, joining AD and AC, the triangles EAD, EAC have the angle E common; also, the angle EAD formed by a tangent and a chord has for its measure half the

E

D

A

C

arc AD (Prop. XX. Bk. III.), and the angle C has the same measure; therefore the angle E A D is equal to the angle C; hence the two triangles are similar (Prop. XXII. Cor.), and give the proportion,

EC: EA:: EA: ED.

2

287. Cor. Hence, EA ECX ED; therefore the square of the tangent is equivalent to the rectangle contained by the whole secant and its external segment.

288. If any angle of a triangle is bisected by a line terminating in the opposite side, the rectangle of the other two sides is equivalent to the square of the bisecting line plus the rectangle of the segments of the third side.

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 D C, being in the same segment (Prop. XVIII. Cor. 1, Bk. III.), the triangles ABE, BCD are similar; hence,

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

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

BDX AÇABX CD+AD X BC.

PROPOSITION XXXIX.-THEOREM.

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

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

D

F

E

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

A G B

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:AB::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 CF 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 two equal parts.

Let A B be a straight line, which it

is required to bisect.

C

E

A

B

XD

From the point A as a centre, with a radius greater than the half of A B, describe an arc of a circle; and *from the point B as a centre, 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.