AC to B. Because AC-DC, the angle ADC=DAC (I. 8. El.); but the angles DAC and DBC are together equal to a right angle (I. 34. El), and therefore equal to ADC and BDC; whence the angle DBC is equal to BDC, and consequently (I. 9. El.) the side DC is equal to BC. The segments AC and BC are thus, each of them, equal to DC, and hence AB itself is double of DC, or is equal to the given straight line.

PROP. VI. THEOR.

If a straight line, drawn from a given point to a straight line given in position, contain a given rectangle, the locus of its point of section will be a given circle.

Let the rectangle AB, AC be given, while the point B and the straight line BD are given in position; the point C will lie in the circumference of a given circle.

ANALYSIS.

B

Draw AD perpendicular to BD, and make the rectangle AD. AE=AB. AC. Since AD is evidently given both in position and magnitude, AE and the point E are given. Join CE. Because ADXAE AB × AC, AD: AB ::AC: AE, and the triangles DAB and CAE, having the sides about the common angle at A proportional, are therefore similar; and consequently the angle ACE is equal to ADB, or a right angle. Whence (III. 26. El.)

D

E

the point E must lie in a semicircle, of which AE, the diameter, is given.

Z

COMPOSITION.

Having drawn the perpendicular AD, make the rectangle AD, AE equal to the given space, and upon the diameter AE describe a circle; this is the locus required. For draw AC and CE. The triangles ABD and AEC are similar, since they have a common angle at A, and those at D and C right angles; wherefore AB: AD :: AE: AC, and ABXAC=ADX AE, that is, equal to the given space.

PROP. VII. THEOR.

If a straight line, containing a given rectangle, be drawn through a given point to the circumference of a given circle, the locus of its point of section will be either a straight line given in position on a given circle, according as it originates, or not, in the given circumference.

Let the rectangle AC, AB be equal to a given space, and the segment AC terminate in a given circumference, the point of origin A may either lie in that circumference

or not.

1. Suppose the given point A lies in the given circumference; the locus of C is the straight line given in position.

10

ANALYSIS.

Draw the diameter AE and make AE× AD=AB×AC;

wherefore the point D is given, and join CE and BD. Because AE × AD = ABX AC, AC: AE :: AD: AB; whence the triangles CAE and DAB, having likewise a common angle at A, are similar. Consequently the angle ADB being thus equal to ACE, is a right angle, and the straight line DB is hence given in position.

COMPOSITION.

E

Having drawn the diameter AE, make the rectangle AE, AD equal to the given space, and erect the perpendicular DB; this is the locus required. For draw ACB and join CE. The right angled triangles ACE and ADB being evidently similar, AC: AE::AD: AB, and ACXAB =AEXAD, or the given space.

2. Suppose that the point A does not lie in the given circumference; then the locus of B is a given circle. 94.

ANALYSIS.

Draw the diameter EAD, and produce CAF to the cir

COMPOSITION.

Having drawn the diameter EAD, make the rectangle AD, AH equal to the given space, and (III. 2.) describe a circle, EBGF, such that a straight line, passing through shall be cut by the circumference in the ratio of AE to AH; this circle is the locus required. For AE: AH :: AF:AB:: AFX AC: AB × AC; wherefore AFX AC: ABXAC: AEXAD: AHX AD, and the first term of this analogy being equal to the third, the second term is equal to the fourth, or AB×AC=AH×AD,that is, equal to the given space.

PROP. VIII. THEOR.

If two straight lines, containing a given rectangle, be drawn from a given point at a given angle; should the one terminate in a straight line given in position, the other will terminate in the circumference of a given circle.

Let the point A, the angle BAC, and the rectangle under its sides, BA, AC, be given; if the direction BD be given, then will the locus of C be a given circle.

1

ANALYSIS.

From A let fall the perpendicular AD upon BD. Draw AE, to contain with AD an angle equal to the given angle, and a rectangle equal to the given space; and join CE.

E

B

அ

Since AD is evidently given in position and magnitude, AE is likewise given in position and magnitude; and the rectangle AD,AE being equal to AB. AC, therefore AD: AB :: AC: AE; but the angle DAE is equal to BAC, and hence DAB is equal to EAC. Wherefore the triangles ABD and AEC, having each an equal angle and its containing sides proportional, are similar; and consequently the angle ACE is equal to the right angle ADB. Whence the locus of C is a circle, having AE for its diameter.

COMPOSITION.

Having let fall the perpendicular AD, draw AE, making the angle DAE equal to the given angle, and the rectangle DA, AE equal to the given space, and on AE, as a diameter, describe a circle; this is the locus required.

For join CE; and the triangles DAB and EAC being right angled at D and C, and having the vertical angles at A equal, are evidently similar, and consequently AD: AB :: AC: AE; and hence the rectangle AB, AC is equal to AD, AE, that is, to the given space.