cutting line is equal to the angle in the alternate segment AFB (iii. Prop. 32); but ▲ BAD = ▲ V (Const.), and therefore the segment AFB contains an angle equal to the given angle V. PROP. XXXIV. PROB. To cut off from a given circle (ABC) a segment which shall contain a given angle (v). B C Draw the straight line DE touching the circle (iii. Prop. 17), and from the point of contact A draw the chord AC, making an angle CAE equal to V (i. Prop. 23), and it will cut off the required segment. For / CAE made by the cutting line AC with the tangent DE is equal to the angle in the alternate segment ABC (iii. Prop. 32); but CAE ZV (Const.), and therefore the angle contained by the segment ABC is equal to the given angle V. = D If two chords (AB, EF) in a circle intersect each other, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other. E If the chords both pass through the centre of the circle, it is evident that their segments, being the rays of the circle, are equal, and that the rectangles contained by them respectively are therefore equal. But if one of the chords EF passing through the centre cuts the other AB not passing through the centre, at right angles; join CA; IC F G and then EF being perpendicular to AB, bisects it (iii. Prop. 3), and AG=GB; and EF being bisected in C and divided unequally in G, EG·GF+ CG3= CF2 (ii. Prop. 5), or CA2, since CA=CF. But CA2=CG2+AG2 (i. Prop. 47); therefore EG GF +CG2=CG2+AG'; and CG' being taken from both, EG GF, or the rectangle contained by the segments of EF, remains F equal to AG2 AG GB, or the rectangle contained by the segments of AB. E G B F But if the chord EF passing through the centre cuts AB obliquely, join CA as before, and draw CD perpendicular to AB, and it will bisect AB, and make AD=DB (iii. Prop. 3). And since AB is divided equally in D, and unequally in G, AG GB+DG-DB2 (ii. Prop. 5); add to both DC2, and then AG·GB+DG2+DC2=DC2+DB2 or AD2, since AD=DB. But DG2+ DC2= CG2 (i. Prop. 47); and in like manner DC2 + AD2 = CA2 = CF2; therefore AG GB +CG-CF2. But also, since EF is divided equally in C and unequally in G, EG·GF+ CG2 CF2; therefore AG GB +CG2 = EG-GF+CG2; and CG2 being taken from both, AG GB, or the rectangle contained by the segments of AB, remains equal to EG GF, or the rectangle contained by the segments of EF. = Lastly, if two chords AB, CD intersect each other, of which neither passes through the centre, it is evident that the rectangles AG-GB and CG GD contained by their respective segments, are both equal to the rectangle EG GF contained by the segments of the chord A which passes through the centre, and that they are therefore equal to each other (Ax. 1). PROP. XXXVI. THEOR. B F If from a point (A) outside of a circle two straight lines be drawn, the one (AB) cutting the circle, the other (AD) touching it, the rectangle contained by the whole secant or cutting line (AB) and its external segment (AF) is equal to the square of the tangent (AD). First, let the secant AB pass through the centre C, and draw CD; and because CD is drawn from the centre to the point of contact, it is perpendicular to the tangent AD (iii. Prop. 18), and ADC is a right angle. But since FB is divided equally in C and produced to A, AB AF+FC2=AC2 (ii. Prop. 6); but FC2=CD2, since CD=FC, and AC2=AD2+CD2 (i. Prop. 47); therefore AB AF+CD2= AD'+CD2; and CD' being taken F B from both, AB AF, or the rectangle contained by the whole secant and its external segment remains equal to AD2 or the square of the tangent. E Next, if the secant AB does not pass through the centre, join CF, and draw CE perpendicular to AB, and it will bisect FB (iii. Prop. 3); and then AB AF+FE2=AE2 (ii. Prop. 6); add to both EC2, and AB AF + FE2 + EC2 = AE2+EC2; but FE2+EC2=CF2 (i. Prop. 47) =CD2; and in like manner AE2+ EC2= AC2=AD2+CD2; therefore AB AF+CD2=AD2+CD2; and CD2 being taken from both, AB AF, or the rectangle contained by the whole secant and its external segment, remains equal to AD2 or the square of the tangent. B If from a point (A) outside of a circle two straight lines be drawn, the one (AB) cutting the circle, the other (AD) meeting it, and if the rectangle contained by the whole cutting line (AB) and its external segment (AG) be equal to the square of the line meeting the circle, the latter (AD) is a tangent to the circle. G From the point A draw to the circle the tangent AF, and join CA, CD and CF. Then the rectangle AB AG = AF2 (iii. Prop. 36); but also AB AG =AD2 (Hyp.); therefore AF-AD2 and AF=AD; and the triangles CAF, CAD having the sides AF=AD, CF=CD, and the base CA common, have the angles contained by the equal sides equal (i. Prop. 8), or ▲ ADC = / AFC: but AFC is a right angle (iii. Prop. 18), and there F fore ADC is also a right angle, and consequently AD is a tangent to the circle (iii. Prop. 16). BOOK IV. DEFINITIONS. 1. A rectilinear figure is said to be inscribed in another rectilinear figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. 2. A rectilinear figure is said to be described about another rectilinear figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. 3. A rectilinear figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle. 4. A rectilinear figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. 5. A circle is said to be inscribed in a rectilinear figure, when the circumference of the circle touches each side of the figure. 6. A circle is said to be described about a rectilinear figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. 7. A straight line is said to be placed in a circle when both its extremities are in the circumference of the circle. PROP. I. PROB. In a given circle (BDC) to place a straight line, equal to a given straight line (A), not greater than the diameter of the circle. Draw BC, the diameter of the given circle; and if it be equal to A, then the problem is solved. But if BC be not equal to A, it must be greater (Hyp.); cut off from it a part, BE,equal to A (i. Prop. 3), and from the centre B, with the interval BE, describe the circle EDF, cutting in D the circle BDC, and join BD, which is the line required. For BD=BE B A (i. Def. 12)=A (Const.), and therefore the straight line BD, placed in the given circle, is equal to the given line A (Ax. 1). PROP. II. PROB. In a given circle (ABC) to inscribe a triangle equiangular to a given triangle (FDE). D of the E B Draw the straight line GH touching the given circle in the point A (iii. Prop. 17); and from the point of contact A, draw the chord AC, making CAH equal to given triangle (i. Prop. 23); and in like manner draw the chord AB, making ZBAG equal to E, and join BC. The triangle ABC is the required triangle. CAH, CAH is ABC For ABC in the alternate segment is equal to made by the chord and tangent (iii. Prop. 32); and equal to D of the given triangle (Const.); therefore, =D; and for a like reason, ▲ ACB = ▲ E; and since Zs ABC, ACB in the one triangle, equal s D, E in the other, the remaining angle BAC in the one must equal the remaining angle F in the other (i. Prop. 32, Cor.). The triangle ABC, therefore, inscribed in the given circle, is equiangular to the given triangle FDE. |