tained by two given lines; find BE a mean proportional between the two given lines, and proceed as in the proposition.

COR. 2. If it be required to produce the line, so that the rectangle contained by the whole line produced and the part produced, may be equal to a given square; take BE equal to a side of the square, and proceed as in the proposition.

(8.) To determine two lines such that the sum of their squares may be equal to a given square, and their rectangle equal to a given rectangle.

A

B

Let AB be equal to a side of the given square. Upon it describe a semicircle ADB; and from B draw BC perpendicular to AB, and equal to a fourth proportional to AB and the sides of the given rectangle. From C draw CD parallel to BA. Join AD, DB; they are the lines required.

Since CB touches the circle at B, the angle CBD is equal to DAB, and the angles DCB, ADB are right angles; the triangles DCB, ADB are equiangular, and AB : AD :: DB : BC,

whence the rectangle AD, DB is equal to the rectangle AB, BC, i. e. to the given rectangle. Also the squares of AD, DB are equal to the square of AB, i. e. to the given square.

(9.) To divide a straight line into two parts, so that

the rectangle contained by the whole and one of the parts may be equal to the square of a given line, which is less than the line to be divided.

Let AB be the given line to be divided.
Upon it describe a semicircle, in which
place the line AC to the given line.
Join CB; and on it describe a semicircle
CDB, cutting AB in D; D is the point required.

Since the angle ACB is in a semicircle, it is a right angle, .. AC touches the circle CDB (Eucl. iii. 16. Cor.); whence the rectangle BA, AD is equal to the square of AC, i. e. to the square of the given line.

(10.) To divide a given line into two such parts that the rectangle contained by the whole line and one of the parts may be (m) times the square of the other part, m being whole or fractional.

Let AB be the given line, and in it produced, take BC=an mth part of AB. On AC describe a semicircle, and from B draw BD perpendicular to AC. Bisect

CB in O; join OD, and take OE=OD; and AB will be divided in E, as required.

On BC describe a semicircle, cutting OD in F; join FE. Then the angle DOE being common to the triangles DOB, EOF, and DO, OB respectively equal to EO, OF, the triangles will be similar and equal, and .. the angle OFE equal to OBD, and .. a right angle; whence FE is a tangent to the circle CFB. Hence the rectangle AB, BC is equal to the square of DB, i. e. to the square of FE, or the rectangle CE, EB.

From each of these equals take away the rectangle CB, BE; and the rectangle AE, CB is equal to the square of BE, . (m) times the rectangle AE, CB, i. e. the rectangle AB, AE is equal to (m) times the square of BE.

(11.) To divide a given line into two such parts that the square of the one shall be equal to the rectangle contained by the other and a given line.

Let AB be the given line to be divided, (see last Fig.) and BC the other given line. Let them be placed so as to be in the same straight line. On AC describe a semicircle and draw the lines, as in the last proposition; and E is the point required.

For the rectangle AE, CB is equal to the square of BE.

(12.) A straight line being given in magnitude and position; to draw to it from a given point, two lines, whose rectangle shall be equal to a given rectangle, and which shall cut off equal segments from the given line.

DE G

Let AB be the given line, and C the given point. Bisect AB in D, and from D draw DO at right angles to AB, and let fall the perpendicular CE. With the centre C, and radius equal to a fourth proportional to 2 CE and the sides of the given rectangle, describe a circle cutting DO in O. Join OC; and with the centre O,

and radius OC, describe a circle CFG, cutting AB in F and G; join CF, CG; they are the lines required.

For (Eucl. vi. C.) the rectangle CF, CG is equal to the rectangle contained by 2 CO and CE, i. e. to the given rectangle. And since AD-DB, and FD=DG, .. AFGB.

(13.) To draw a straight line which shall touch a given circle, and make with a given line, an angle equal to a given angle.

E

Let AB be the given line, and O the centre of the given circle. From any point in the given line, draw AC making with it an angle equal to the given angle; from O draw OD perpendicular to AC, and through the point E where it meets the circle, draw EF parallel to DA; EF is the line required.

A

F

B

For being parallel to AC it is perpendicular to OD, and a tangent to the circle; and the angle EFB = DAB the given angle.

=

(14.) Through a given point to draw a line terminating in two lines given in position, so that the rectangle contained by the two parts may be equal to a given rectangle.

Let AB, CD be the lines given in position, E the given point; from E draw EF perpendicular to AB, and

and produce FE to G, so that the rectangle FE, EG may be equal to the given rectangle. On EG describe a circle, cutting CD in H. Join HE, and produce it to A; AH is the line required.

C

F

II

Join GH. The angle GEH is equal to AEF, and the angles GHE, AFE are right angles, .. the triangles GEH, AEF are equiangular, and

EH EG: EF: EA,

whence the rectangle AE, EH is equal to the rectangle EG, EF, i. e. to the given rectangle.

(15.) From a given point to draw a line cutting two given parallel lines, so that the difference of its segments may be equal to a given line.

Let AB, CD be the given parallels, and P the given point. From P draw any line PB, meeting the given lines in B and E. Make EF = EP, and draw FG

parallel to AB. With any point O as centre, and radius equal to the given line, describe a circle cutting GF in H. Join OH, and draw PGA parallel to it. PGA will be the line required.

Since PE is equal to EF, .. (Eucl. vi. 2.) PI=IG; and AG is equal to the difference of AI and IP, the segments of PA; and AG=OH= the given line.