Let the triangle ABC have the angle BAC bisected by the straight line A D terminating in the opposite side BC; then the rectangle BAX AC is equivalent to the square of AD plus the rectangle BD X D C. Describe a circle through the three points A, B, C ; produce A D till A meets the circumference at E, and join CE.

=

B

D

A

The triangles BAD, EAC have, by hypothesis, the angle BAD equal to the angle EA C; also the angle B equal to the angle E, being measured by half of the same arc AC (Prop. XVIII. Cor. 1, Bk. III.); these triangles are therefore similar (Prop. XXII. Cor.), and their homologous sides give the proportion,

BA: AE:: AD: AC;

hence,

BAXACA EX AD.

But A E is equal to AD + D E, and multiplying each of these equals by AD, we have,

AEX AD AD2+ ADX DE;

C

now, AD × DE is equivalent to BD X DC (Prop. XXXIII. Cor.); hence

BAXACAD2+ BD X DC.

PROPOSITION XXXVII.-THEOREM.

289. The rectangle contained by any two sides of a triangle is equivalent to the rectangle contained by the diameter of the circumscribed circle and the perpendicular drawn to the third side from the vertex of the opposite angle.

In any triangle A B C, let AD be drawn perpendicular to BC; and let E C be the diameter of the circle circum

scribed about the triangle; then will ABX AC be equivalent to ADX CE.

For, joining AE, the angle EAC is a right angle, being inscribed in a semicircle (Prop. XVIII. Cor. 2, Bk. III.); and the angles B and E are equal, being measured by half of the same arc, A C (Prop. XVIII. Cor. 1, Bk. III.); hence the two rightangled triangles are similar (Prop. XXII. Cor.), and give the proportion AB:CE::AD: A C; hence

E

A

ABX ACCEX AD.

290. Cor. If these equals be multiplied by BC, we shall have

A

Ꭰ

ABX ACX BC= CEX ADX BC.

But ADX BC is double the area of the triangle (Prop. VI.); therefore the product of the three sides of a triangle is equal to its area multiplied by twice the diameter of the circumscribed circle.

E

PROPOSITION XXXVIII.-THEOREM.

291. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle is equivalent to the sum of the two rectangles of the opposite sides.

Let ABCD be any quadrilateral inscribed in a circle, and A C, BD its diagonals; then the rectangle ACX BD is equivalent to the sum of the two rectangles ABX CD, ADX BC.

For, draw BE, making the angle ABE equal to the angle CBD; to each of these equals add the angle EBD, and we shall have the angle ABD equal to the angle EBC; and the

B

C

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, A
AD: BD::CE: BC;

and, consequently,

=

=

ADX BC BDX CE.

Again, since the angle ABE is equal to the angle CBD, and the angle B A E is equal to the angle B D C, 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 BDXAE+BD × CE=BD (AE÷CE)=BD × AC, we have

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

B

Ꭰ

E

D

BDX ACAB× CD+AD × BC.

PROPOSITION XXXIX.-THEOREM.

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

F

C

E

To find a common measure, if there be one, we must apply A B, 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 AC,

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 AF 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 AB, or its equal CF, with AE; which will result, as before, in leaving a remainder AF; 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.

XC

A.

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.

E

B

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.