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f. 28. Dat position cut one another is given f. and the straight line DB which 8.33 Dat is at right angles to AB is given 8 in position, and AB is given in
position, therefore f the point B is given. and the points A, D are k. 29. Dat given, wherefore the straight lines AB, BD are given. and the i. 2. Dat. ratio of AB to BC is given, and therefore i BC is given.
The Composition. Let the given ratio of FG to GH be that which AB is required to have to BC, and let HK be the given straight line which is to be taken from BC, and let the ratio which the remainder is required to have to BD be the given ratio of HG to GL, and place GL at right angles to FH, and join LF, LH. Next, as HG is to
H K GF, so make HK to AE; produce AE to N so that AN be the straight line to the square of which the sum of the squares of AB, BD is required to be equal; and make the angle NED equal to the angle GFL. from the center A at the distance AN describe a circle, and let its circumference meet ED in D, and draw DB perpendicular to AN, and DM making the angle BDM equal to the angle GLH. lastly, produce BM to C so that MC be equal to HK. then is AB the first, BC the second and BD the third of the straight lines that were to be found.
For the triangles EBD, FGL, as also DBM, LGH being equiangular, as EB to BD, so is FG to GL; and as DB to BM, so is
LG to GH; therefore, ex aequali, as EB to BM, so is (FG to k. 12. 5. GH, and so is) AE to HK or MC; wherefore * AB is to BC, as
AE to HK, that is, as FG to GH, that is, in the given ratio. and from the straight line BC taking MC which is equal to the given
straight line HK, the remainder BM has to BD the given ratio of d. 47. 1. HG to GL. and the sum of the squares of AB, BD is equal d to
the square of AD or AN which is the given space. Q. E. D.
I believe it would be in vain to try to deduce the preceding Construction from an Algebraical Solution of the Problem.
PLANE AND SPHERICAL
PRINTED BY J. & M. ROBERTSON,
LEMMA I. Fig. 1.
center, and with any distance BA, a circle be described, meeting BA, BC, the straight lines including the angle ABC in A, C; the angle ABC will be to four right angles, as the arch AC to the whole circumference.
Produce AB till it meet the circle again in F, and through B draw DE perpendicular to AB, meeting the circle in D, E.
By 33. 6. Elem. the angle ABC is to a right angle ABD, as the arch AC to the arch AD; and quadrupling the consequents, the angle ABC will be to four right angles, as the arch AC to four times the arch AD, or to the whole circumference.
LEMMA II. Fig. 2. ET ABC be a plane rectilineal angle as before ; about B as be described meeting BA, BC in D, E, A, C; the arch AC will be to the whole circumference of which it is an arch, as the arch DE is to the whole circumference of which it is an arch.
By Lemma 1. the arch AC is to the whole circumference of which it is an arch, as the angle ABC is to four right angles; and by the same Lemma 1. the arch DE is to the whole circumference of which it is an arch, as the angle ABC is to four right angles; therefore the arch AC is to the whole circumference of which it is an arch, as the arch DE to the whole circumference of which it is an arch. DEFINITIONS. Fig. 3.
ET ABC be a plane rectilineal angle; if about B as a center,
with BA any distance, a circle ACF be described meeting BA, BC, in A, C; the arch AC is called the measure of the angle ABC.
II. The circumference of a circle is supposed to be divided into 360
equal parts called degrees, and each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds, &c. And as many degrees, minutes, seconds, &c. as are contained in any arch, of so many degrees, mi.
nutes, seconds, &c. is the angle, of which that arch is the
measure, said to be. COR. Whatever be the radius of the circle of which the mea
fure of a given angle is an arch, that arch will contain the
CBF, which, together with ABC, is equal to two right angles,
the arch AC, perpendicular upon the diameter passing through
of the angle ABC, of which it is the measure.
mity of the arch AC between the fine CD, and that extremity
the arch AC, and meeting the diameter BC passing through
tangent AE, is called the Secant of the arch AC, or angle
angle ABC, are likewise the fine, tangent, and secant of its
supplement CBF. It is manifest from Def. 4. that CD is the fine of the angle CBF.
Let CB be produced till it meet the circle again in G; and it is manifest that AE is the tangent, and BE the fecant, of the
angle ABG or EBF, from Def. 6. 7. Cor. to Def. 4. 5. 6. 7. The fine, verfed fine, tangent, and
fecant, of any arch which is the measure of any given angle ABC, is to the fine, versed line, tangent, and fecant, of any