THEOREM XXIV.
PL. 1. fig. 8.

Let DHB be a quadrant of a circle described by the radius CB; HB an arc of it, and DH its complement; HL or FC the sine, FH or CL its cosine, BK its tangent, DI its cotangent; CK its secant, and CI its cosecant. Fig. 8.

1. The cosine of an arc is to the sine as the radius is to the

tangent.

2. The radius is to the tangent of an arc as the cosine of it is to the sine.

3. The sine of an arc is to its cosine as the radius to its cotangent.

4. Or the radius is to the cotangent of an arc as its sine to its co-sine.

5. The cotangent of an arc is to the radius as the radius to the tangent.

6. The cosine of an arc is to the radius as the radius is to the secant.

7. The sine of an arc is to the radius as the tangent is to the secant.

The triangles CLH and CBK being similar (by theo. 16), 1. CL: LH:: CB: BK.

2. Or, CB: BK:: CL: LH.

The triangles CFH and CDI being similar,

3. CF (or LH): FH :: CD: DI.

4. CD: DI:: CF (or LH): FH.

The triangles CDI and CBK are similar; for the angle CID=KCB, being alternate ones (by part 2, theo. 3), the lines CB and DI being parallel, the angle CDI=CBK being both right, and consequently the angle DCI=CKB, wherefore,

5. DI:CD::CB:BK.

And again, making use of the similar triangle CLH and CBK,

6. CL: CB:: CH: CK. 7. HL: CH:: BK: CK.

GEOMETRICAL PROBLEMS.

PROBLEM I.

PL. 2. fig. 7.

To make a triangle of three given right lines BO, LB, LO, of which ang two must be greater than the third.

Lay BL from B to L; from B with the line BO describe an arc, and from L with LO describe another arc; from O, the intersecting point of those arcs, draw BO and OL, and BOL is the triangle required.

This is manifest from the construction.

PROBLEM II.
PL. 2. fig. 8.

At a point B in a given right line BC, to make an angle equal to a given angle A.

Draw any right line ED to form a triangle, as EAD, take BF -AD, and upon BF make the triangle BFG, whose side BG= AE, and GF=ED (by the last), then also the angle B=A; if we suppose one triangle be laid on the other, the sides will mutually agree with each other, and therefore be equal; if we consider these two triangles to be made of the same three given lines, they are manifestly one and the same triangle. Otherwise,

for

Upon the centres A and B, at any distance, let two arcs DE, FG, be described; make the arc FG-DE, and through B and G draw the line BG, and it is done.

For since the chords ED, GF are equal, the angles A and B are also equal, as before (by def. 17).

PROBLEM III.
PL. 2. fig. 9.

To bisect or divide into two equal parts any given right-lined angle BAC. In the lines AB and AC, from the point A, set off equal distances AE=AD; then, with any distance more than the half of DE, describe two arcs to cut each other in some point F; and the right line AF, joining the points A and F, will bisect the given angle BAC.

For if DF and FE be drawn, the triangles ADF, AEF are equilateral to each other, viz: AD=AE, DF=FE, and AF common, wherefore DAF=EAF, as before.

PROBLEM IV.

PL. 2. fig. 10.

To bisect a right line AB.

With any distance more than half the line from A and B, describe two circles CFD, CGD, cutting each other in the points C and D; draw CD intersecting AB in E, then AE=EB.

For, if AC, AD, BC, BD be drawn, the triangles ACD, BCD will be mutually equilateral, and consequently the angle ACE=BCE; therefore the triangle ACE, BCE, having AC BC, CE common, and the angle ACE=BCE; (by theo. 6) the base AE=the base_BE.

Cor. Hence it is manifest that CD not only bisects AB, but is perpendicular to it (by def. 10).

PROBLEM V.

PL. 2. fig. 11.

On a given point A, în a right line EF, to erect a perpendicular. From the point A lay off on each side the equal distances AC, AD; and from C and D as centres, with any interval greater than AC or AD, describe two arcs intersecting each other in B; from A to B draw the line AB, and it will be the perpendicular required.

For let CB and BD be drawn, then the triangles CAB, DAB will be mutually equilateral and equiangular, so CAB=DAB, a right angle (by def. 10).

PROBLEM VI.

PL. 2. fig. 12.

To raise a perpendicular on the end B of a right line AB.

From any point D not in the line AB, with the distance from D to B, let a circle be described cutting AB in E; draw from E through D the right line EDC, cutting the periphery in C, and join CB, and that is the perpendicular required.

EBC being a semicircle, the angle EBC will be a right angle (by cor. 5, theo. 7).

ROBLEM VII.

PL. 2. fig. 13.

From a given point A, to let fall a perpendicular upon a given right line BC.

From any point D, in the given line, take the distance to the given point A, and with it describe a circle AGE, make GE=

AG, join the points A and E by the line AFE, and AF will be the perpendicular required..

Let DA, DE be drawn, the angle ADF=FDE, DA=DE, being radii of the same circle, and DF common; therefore (by theo. 6) the angle DFA=DFE, and FA a perpendicular. (By def. 10.)

PROBLEM VIII.

PL. 2. fig. 14.

Through a given point A to draw a right line AB, parallel to a given right line CD.

From the point A to any point F in the line CD draw the line AF; with the interval FA, and one foot of the compasses in F, describe the arc AE, and with the like interval and one foot in A describe the arc BF, making BF-AE; through A and B draw the line AB, and it will be parallel to CD. By prob. 2, The angle BAF-AFE, and by theo. 11, BA and CD are parallel.

'PROBLEM IX.
PL. 1. fig. 17.

Upon a given line AB to describe a square ABCD.

Make BC perpendicular and equal to AB, and from A and C, with the line AB or BC, let two arcs be described, cutting each other in D; from whence to A and C let the lines AD, DC be drawn; so is ABCD the square required.

For all the sides are equal by construction; therefore the triangles ADC and BAC are mutually equilateral and equiangular, and ABCD is an equilateral parallelogram, whose angles are right. For B being right, D is also right, and DAC, DCA, BAC, ACB, each half a right angle (by lemma preceding theo. 7, and cor. 2, theo. 5), whence DAB and BCD will each be a right angle, and (by def. 43) ABCD is a

square.

SCHOLIUM.

By the same method a rectangle or oblong may be described, the sides thereof being given.

PROBLEM X.
PL. 2. fig. 15.

To divide a given right line AB into any proposed number of equal parts.

Draw the indefinite right line AP, making any angle with

AB, also draw BQ parallel to AP, in each of which let there be taken as many equal parts AM, MN, &c. Bo, on, &c. as you would have AB divided into; then draw Mm, Nn, &c. intersecting AB in E, F, &e. and it is done.

For MN and mn being equal and parallel, FN will be parallel to EM, and in the same manner GO to FN (by theo. 12); therefore AM, MN, NO, being all equal by construction, it is plain (from theo. 20) that AE, EF, FG, &c. will likewise be equal.

PROBLEM XI.
PL. 2. fig. 16.

To find a third proportional to two given right lines A and B. Draw two indefinite blank lines CE, CD anywise to make any angle. Lay the line A from C to F, and the line B from C to G, and draw the line FG; lay again the line A from C to H, and through H draw HI parallel to FG (by prob. 8), so is CI the third proportional required.

For, by cor. 1, theo. 20, CG: CH:: CF: CI.
Or, B: A :: A: CI.

PROBLEM XII.

PL. 2. fig. 17.

Three right lines A, B, C given, to find a fourth proportional. Having made an angle DEF anywise, by two indefinite blank right lines ED, EF, as before; lay the line A from E to G, the line B from E to I, and draw the line IG; lay the line C from E to H, and (by prob. 8) draw HK parallel thereto, so will EK be the fourth proportional required.

For, by cor. 1, theo. 20, EG : EI :: EH : EK.
Or, A: B:: C: EK.

PROBLEM XIII.
PL. 3. fig./1.

Two right lines A and B given, to find a mean proportional. Draw an indefinite straight line, on which place AB=A and BC=B; bisect AC (by prob. 4) in E, and describe the semicircle ADC, and from the point B erect the perpendicular BD (by prob. 5), then BD is a mean proportional.

For if the lines AD, DC be drawn, the angle ADC is a right angle (by cor. 5, theo. 7), being an angle in a semicircle.

The angles ABD, DBC are right ones (by def. 10), the line BD being a perpendicular; wherefore the triangles ABD,