or IN, and KM IK, therefore KL is equal to KM2 + KN, or the square of the longest radius, of the said circular sections, is equal to the sum of the And besquares of the two others.

cause circles are to each other as the squares of their diameters, or of their radii, therefore the circle described by KL is equal to both the circles described by Kм and KN; or the section of the cylinder, is equal to both the corresponding sections of the sphere and cone. And as this is always the case in every parallel position of KL, it follows, that the cylinder, EB, which is composed of all the former sections, is equal to the hemisphere EFG and cone IAB, which are composed of all the latter sections.

But the cone IAB is a third part of the cylinder EB (cor. 2, th. 115); consequently the hemisphere EFG is equal to the remaining two-thirds; or the whole sphere EFGH equal to two-thirds of the whole cylinder ABCD. Q. E. D.

Corol 1. A cone, hemisphere, and cylinder of the same base and altitude, are to each other as the numbers 1, 2, 3.

Corol. 2. All spheres are to each other as the cubes of their diameters; all these being like parts of their circumscribing cylinders.

Corol. 3. From the foregoing demonstration it also appears, that the spherical zone or frustrum EGNP, is equal to the difference between the cylinder EGLO and the cone IMQ, all of the same common height IK. And that the spherical segment PFN, is equal to the difference between the cylinder ABLO and the conic frustrum AQMB, all of the same common altitude FK,

PROBLEMS.

PROBLEM S.

PROBLEM I.

To Bisect a Line AB; that is, to divide it into two Equal

Parts.

FROM the two centres A and B, with any equal radii, describe arcs of circles, intersecting each other in c and D, and draw the line CD, which will bisect the given line AB in the point E.

For, draw the radii AC, BC, AD, BD. Then, because all these four radií are equal, and the side CD common, the two

B

triangles ACD, BCD, are mutually equilateral: consequently they are also mutually equiangular (th. 5), and have the angle ACE equal to the angle BCE.

Hence, the two triangles ACE, BCE, having the two sides. AC, CE, equal to the two sides BC, CE, and their contained angles equal, are identical (th. 1), and therefore have the side AE equal to EB. Q. E. D.

PROBLEM II.

To Bisect an Angle BAC.

FROM the centre a, with any radius, describe an arc, cutting off the equal lines AD, AE; and from the two centres D, E, with the same radius, describe arcs intersecting in F; then draw AF, which will bisect the angle A as required. For, join DF, EF.

Then the two tri

angles ADF, AEF, having the two sides

D

B

A

AD, DF, equal to the two AE, EF (being equal radii), and the side AF common, they are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angle BAF equal to the angle CAF.

Scholium. In the same manner is an arc of a circle bisccted.

VOL. I.

Z z

PROBLEM

PROBLEM III.

At a Given Point c, in a Line AB, to Erect a Perpendicular,

FROM the given point c, with any radius, cut off any equal parts CD, CE, of the given line; and, from the two centres D and E, with any one radius, describe arcs intersecting in F; then join CF, which will be perpendicular as required.

CEB

Then the two

For, draw the two equal radii DF, EF. triangles CDF, CEF having the two sides CD, DF, equal to the two CE, EF, and CF common, are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the two adjacent angles at c equal to each other; there fore the line cr is perpendicular to AB (def. 11).

Otherwise.

When the Given Point c is near the End of the line.

FROM any point D, assumed above the line, as a centre, through the given point o describe a circle, cutting the given line at E; and through E and the centre D, draw the diameter EDF; then join CF, which will be the perpendicular required.

D

CB

For the angle at c, being an angle in a semicircle, is a right angle, and therefore the line CF is a perpendicular (by def. 15).

PROBLEM IV.

From a Given point A to let fall a Perpendicular on a given Line sc.

FROM the given point a as a centre, with any convenient radius, describe an arc, cutting the given line at the two points D and E; and from the two centres D, E with any radius, describe two arcs, intersecting at F; then draw AGF, which will be perpendicular to BC as required.

For, draw the equal radii AD, AE, and

DF, EF.

A

Then the two triangles ADF, AEF, having the two sides AD, DF, equal to the two AE, EF, and AF common, are

mutually

mutually equilateral; consequently they are also mutually equiangular (th 5), and have the angle DAG equal the angle EAG Hence then, the two triangles ADG, AEG, having the two sides AD, AG, equal to the two AE, AG, and their included angles equal, are therefore equiangular (th. 1), and have the angles at & equal; consequently AG is perpendicular to Bc (def 11).

Otherwise.

When the Given Point is nearly Opposite the end of the Line.

FROM any point D, in the given line BC, as a centre, describe the arc of a circle through the given point A, cutting BC in B, and from the centre E, with the radius EA, describe another arc, cutting the former in F; then draw AGF, which will be perpendicular to BC as required.

BD

GE

F

For, draw the equal radii DA DF, and EA EF. Then the two triangles DAE, DFE, will be mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angles at D equal. Hence, the two triangles DAG, DFG, having the two sides DA, DG, equal to the two DF, DG, and the included angles at D equal, have also the angles at G equal (th. 1); consequently those angles at & are right angles, and the line AG is perpendicular to DG.

PROBLEM V.

At a Given Point A, in a Line AB, to make an Angle Equa! to a Given Angle c.

FROM the centres A and c, with any one radius, describe the arcs DE, FG. Then, with radius DE, and centre F, describe an arc, cutting FG in G Through G draw the line AG, and it will form the angle required.

E

G

A

FB

For, conceive the equal lines or radii, DE, FG, to be drawn. Then the two triangles CDE, AFG, being mutually equilateral, are mutually equiangular (th. 5), and have the angle at a equal to the angle o.

PROBLEM

PROBLEM VI

Through a Given Point A, to draw a Line Parallel to a Given Line BC.

EA

F

FROM the given point A draw a line AD to any point in the given line BC. Then draw the line EAF making the angle at a equal to the angle at D (by prob 5); so shall EF be parallel to BC as required. For, the angle D being equal to the alternate angle A, the lines BC, EF, are parallel, by th. 13.

PROBLEM VII.

B

DC

To Divide a Line AB into any proposed Number of Equal

Parts.

DRAW any other line Ac, forming any angle with the given line AB; on which set off as many of any equal parts, AD, DE, EF, FC, as the line AB is to be divided into. Join BC; parallel to which draw the other lines FG, EH, DI: then these will divide

AB in the manner as required-For those parallel lincs divide both the sides AB, AC, proportionally, by th. 82.

PROBLEM VIII.

To find a Third Proportional to Two given Lines AB, AC.

PLACE the two given lines AB, AC, forming any angle at A ; and in AB take also AD equal to AC. Join BC, and draw DE parallel to it; so will AE be the third proportional sought.

For, because of the parallels BC. DE,

the two lines AB, AC, are cut propor

A

-B

A

C

E

A

DB

tionally (th. 82); so that AB AC :: AD or AC AE; there fore AE is the third proportional to AB, AC.

PROBLEM IX.

To find a Fourth Proportional to three Lines AB, AC, AD.

PLACE two of the given lines AB. AC, making any angle at A; also place AD on AB. Join BC; and parallel

to