Describe BC, the square of AB (XLVI, Book I.); to AC apply a parallelogram equal to B C and exceeding by C a figure A D similar to B C (XXIX); since AD is similar to BC it is a square; since BC and C D are equal, if C E be taken away from both, B F and A D are equal, and they are equiangular, therefore EF is A to ED as EA to E B (XIV); but E F and ED are equal to A B and A E, therefore A B is to AE as A E to E B.

Otherwise thus:

E B

D

Divide A B in E so that the rectangle under A B and E B shall be equal to the square of A E (XI, Book II.), and A B is to AE as AE to EB (XVII), therefore A B is cut in extreme and mean

ratio.

(649) If the lesser segment be taken upon the greater, the greater will be cut in extreme and mean ratio (281); and by continuing this process a series of lines will be found in continued proportion, in which the common ratio is that of the segments of a line divided in extreme and mean ratio.

The problem to divide a line in extreme and mean ratio is only a particular case of the following more general one.

*

* (650) To divide a line so that the rectangle under the whole line and one pari shall bear a given ratio (m: n) to the square of the other part.

I

n

m

Let any line E F be taken as diameter, and let a circle be described. Take a third proportional to m and n, and upon the tangent at E take EG, a fourth proportional to m, l, and E F. Then draw G H through the centre C, and cut A B at D so that AD: DB=HI: IG, and it will be cut as required.

H

C

B

A B B

For EF: EG=m: l, that is, in the duplicate ratio of m:n, the squares of E F and E G are as m: n. But the square of E G is equal to the rectangle H G x G I. Therefore H G x GI: the square of HI=m: n. Hence H G is cut as required at I, and A B is similarly cut at D.

*

*

*(651) In the solution of this problem we have assumed that if HI:IG=AD: D B the rectangle H G × GI: the square of HI≥ ABX BD: the square of AD. This may be easily proved. Since HI: IG=AD: DB, . H G : GI= AB: BD, . the rectangles HGX GI and A B x BD are similar. Also the squares of H I and AD are similar. But HI : I G = A D : D B, ·.· HI : AD = IG : D B. Therefore the similar squares on HI and A D are as the similar rectangles HG × GI, AB × BD on IG and B D. In the same manner the converse of this may be proved, scil. if H G x GI: the square of H IAB × BD: the square of AD, then HI:IG= AD: DB.

*

** (652) Hence if two lines be cut in extreme and mean ratio they are cut similarly, and if a line be cut in extreme and mean ratio, any line cut similarly will be also cut in extreme and mean ratio.

*** (653) If the perpendicular B C in a right angled triangle divide the hypotenuse A D in extreme and mean ratio, the lesser side CD is equal to the alternate segment A B, and vice versâ.

=

А В.

A

B D

For (553) DA × B D = the square of D C, but (hyp.) it also equals the square of A B, ·.· C D : Also if C D = A B, ·.· DA X BD the square A B, .. A D is cut in extreme and mean ratio at B. * * (654) The three sides of such a right angled triangle are in continued proportion, and vice versa. For AD × A B = the square of AC (553), but CD AB (hyp.), . A D× CD = the square of CA. Also if A D x CD the square of A C, it also = A D × AB, CDA B.

*

** (655) Hence on a given hypotenuse a right angled triangle may be constructed whose sides are in continued proportion by dividing the given hypotenuse in extreme and mean ratio, describing a semicircle on it, and drawing a perpendicular to meet the semicircle, &c.

(656) If any similar rectilinear figures be similarly described on the sides of a right angled triangle (B A C), the figure described on the side (B C) subtending the right angle is equal to the sum of the figures on the other side.

B D

C

From the right angle draw a perpendicular A D to the opposite side; BC is to CA as CA to CD (550), therefore the figure upon BC is to the similar figure upon CA as BC to CD (XX), but the figure upon B C is to the similar figure upon BA as B C to BD (XX). Hence the sum of the segments B D and CD is to the hypotenuse B C as the sum of the figures on the sides is to the figure on the hypotenuse. But the sum of the segments is equal to the hypotenuse, and therefore the sum of the figures on the sides is equal to the figure on the hypotenuse.

This proposition might be more immediately deduced from the twentysecond proposition of this book and the forty-seventh of the first. Any similar figures on the hypotenuse and sides are as the squares of these lines (XXII); but the sum of the squares of the sides is equal to the square of the hypotenuse, and therefore the sum of any similar figures on the sides is equal to the figure on the hypotenuse.

(656) If two triangles (A B C, CD E) have two sides proportional (A B to B C as CD to DE), and

be so placed at an angle that the homologous sides are parallel, the remaining sides (A C and C E) form one right line.

A

B

с

E

Because A B and C D are parallel, the alternate angles B and BCD are equal (XXIX, Book I.), and also since C B and ED are parallel, the angles D and BCD are equal (XXIX, Book I.), therefore B and D are equal; and since the sides about these angles are proportional (hyp.), the triangles ABC and CDE are equiangular (VI), therefore the angles A C B and C ED are equal; but B CD is equal to CD E, and if DC E be added, A CD and DCE are together equal to CE D, EDC, and D CE; therefore A CD and D C E are equal to two right angles (XXXII, Book I.), and therefore A C and C E form one right line (XIV, Book I.).

B

In the enunciation of this proposition, it should be stated that the proportional sides of the triangles which are not homologous should form the angle at which they are joined, for otherwise the remaining sides might not lie in the same right line. The triangles might be placed as in the annexed figure where A B : B C = CD: DE, and A

the sides A B and C D, as also B C and D E, are respectively parallel, but the angles A B C and C D E are supplemental, and A C and CE are obviously not in the same right line.

(657) In equal circles, angles, whether at the centres (B G C, E HF) or circumferences (BAC, EDF), have the same ratio which the arcs on which they stand have to one another: so also have the sectors (B G C, E B F.)

Take any number of arcs CK, K L, each equal to B C, and any number whatever F M, M N, each equal to E F: and join GK, GL, H M, H N. Because the arcs B C, C K, K L, are all equal, the angles B G C, CG K, K G L are also all

H

K

ས

M

(XXVII, Book III.) equal: therefore what multiple soever the arc BL is of the arc B C, the same multiple is the angle B G L of the angle BGC: for the same reason, whatever multiple the arc E N is of the arc EF, the same multiple is the angle E HN of the angle EHF: and if the arc B L be equal to the arc EN, the angle BGL is also equal (XXVII, Book III.) to the angle EH N; and if the arc BL be greater than EN, likewise the angle B G L is greater than EHN; and if less, less: therefore as the arc B C to the arc EF, so (Def. V. Book V.) is the angle BGC to the angle EHF: but as the angle BGC is to the angle EHF, so is (XV, Book V.) the angle BAC to the angle EDF: for each is double (XX, Book III.) of each; therefore as the arc B C is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle E DF.

A

D

M

F

Also, as the arc B C to E F, so shall the sector B G C be to the sector EHF. Join B C, CK, and in the arcs BC, CK take any points X, O, and join B X, X C, CO, OK: then, because in the triangles G B C, GC K the two sides B G, G C are equal to the two CG, GK, each to each, and that they contain equal angles, the base B C is equal (IV, Book I.) to the base C K, and the triangle G B C to the triangle G C K; and because the arc B C is equal to the arc C K, the remaining part of the circumference of the circle A B C is equal to the remaining part of the circumference of the same circle: therefore the angle B XC is equal (XXVII, Book III.) to the angle COK; and the segment B XC is therefore similar to the segment C OK: and they are upon equal straight lines, B C, C K, and are equal (XXIV, Book III.); therefore the segment BX C is equal to the segment COK: and the triangle B G C was proved to be equal to the triangle C G K; therefore the sector BGC is equal to the sector CGK: for the same reason, the sector K GL is equal to each of the sectors BGC, CG K: in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another: therefore, what multiple soever the arc BL is of B C, the same multiple is the sector B G L of the sector B G C and for the same reason, whatever multiple the arc EN is of E F, the same multiple is the sector E H N of the sector EHF: and if the arc BL be equal to E N, the sector B G L is equal to the sector E HN; and if the arc BL be greater than EN, the sector B G L is greater than the sector EHN; and if less, less: therefore as (Def. V. Book V.) the arc BC is to the arc EF, so is the sector B G C to the sector EHF.

:

By this proposition we are entitled to assume arcs referred to the same radius as measures of angles, and vice versâ.

(659) Every arc is to a quadrant of the same circle as the corresponding central angle is to a right angle, and it is to the whole circumference as the same angle to four right angles.

(660) Similar arcs of different circles being those which are proportional to their circumferences must subtend equal angles at the centre and circumference.

(661) Also in different circles arcs which subtend equal angles at the centre or circumference are similar.

Hence similar segments are contained by similar arcs, and vice versâ. (662) The arcs of unequal circles are in a ratio compounded of their central angles and their radii.

Let A, A' be the arcs, R, R' the radii, and a, a' the angles. With a radius equal to R describe an angle equal to a', and let the subtending arc be m.

Since the arcs A and m have equal radii, :: A : m = a : a', and since m and A' have equal central angles they are as their radii (633), m: A'R R'. But A: A' is a ratio compounded of the ratios A: m and m : A', or of the equivalent ratios a: a' and R: R'. (663) Central angles are in a ratio compounded of the direct ratio of their arcs, and the inverse ratio of their radii.

For by (662) we have A : A' =

{

a: a'. R: R'.

Let each of these equal ratios be compounded with the ratio R': R, and we have