Because FA is a tangent to the circle, and AC cuts (1)Prop.32 it, the angle in the segment ABC is equal to FAC B. 3. (1), and therefore equal to the given angle V (2). (2) Constr. PROP. XXXV. THEOR. If two right lines (AB and CD) within a circle cut Fig. 48,49 . one another, the rectangle under the segments (AE and $ 50. EB) of one of them is equal to the rectangle under the segments ( CE and ED) of the other. 1. If the given right lines pass through the centre, they are bisected in the point of intersection, therefore the rectangles under their segments are the squares of their halves, and therefore are equal. 2. Let one of the given lines (DC) pass through the centre, and the other AB not; draw OA and OB. (1) Cor. The rectangle AEB is equal to the difference between Prop.6.3.2 the squares of OE and of OA (1), that is, to the dif- (2) Prop.5. ference between the squares of OE and of OC or to the rectangle DEC (2). 3. Let neither of the given lines pass through the centre; draw through their intersection a diameter FG, and the rectangle under FE and EG is equal to the rectangle under DE and EC and also to the rect (3) Part angle under BE and EA (3), therefore the rectangle prec. under DE and EC is equal to the rectangle under BE (4) 44. 1. and EA (4). B. 2. PROP, XXXVI. THEOR. If from a point (B) without a circle two right lines Fig. 51 & be drawn to it, one of which (BF) is a tangent to the 52. circle, and the other (BC) cuts it, the rectangle under the whole secant (BC) and the external segment (BO) is equal to the square of the tangent (BF). 1. Let BC pass through the centre; draw AF from Fig. 51. the centre to the point of contact; the square of BF B. 2. B. 2. is equal to the difference between the squares of BA (1) Prop.47 and of AF (1), that is, to the difference between the (2) Prop.6. squares of BA and of A0, or to the rectangle under CB and BO (2). Fig. 52. 2. If BC does not pass through the centre, draw (1) Cor.P.6 AO and AC. The rectangle under CB and BO is equal to the difference between the squares of AB and (2) Prop.47 of AO(1), that is to the difference between the squares of AB and AF or to the square of BF (2). Cor. 1. Hence, if from any point without a circle two right lines be drawn cutting the circle, the rectangles under them and their external segments are equal, for each of the rectangles is equal to the square of the tangent. Cor. 2. If from the same point two lines be drawn to a circle, which are tangents to it, they are equal; for their squares are equal to the same rectangle. PROP. XXXVII. THEOR. Fig. 53. If from a point (B) without a circle two right lines le drawn, one (BC) cutting the circle, the other (BF) meeting it, and if the rectangle under the secant and its external segment be equal to the square of the line which meets the circle, the line (BF) which meets is a tangent. B. 3. Draw from the point B the line BQ a tangent to the circle, and draw EF and EQ. The square of BQ is equal to the rectangle under (1) Prop.36 BC and BO (1), but the square of BF is also equal to (2) Hypoth. the rectangle under BC and BO (2), therefore the (3) Cor. 2. squares of BF and BQ are equal, and therefore the P. 46.B.1. fines themselves (3); then in the triangles EFB and (4) Prop.8. EQB the sides EF and FB are equal to the sides EQ (5) Prop.18 and QB, and the side EB is common, therefore the angle EFB is equal to EQB (4), but EQB is a right (6) Prop.16 angle (5), therefore ÉFB is a right angle, and therefore B. 3. the right line BF is a tangent to the circle (6). B. 1. B.3. THE ELEMENTS OF EUCLID. BOOK IV. DEFINITIONS. 1. A rectilineal figure is said to be inscribed in a Plate 4. circle, when the vertex of each angle of the figure is Fig.1.1.4. Ses N. in the circumference of the circle. 2. A rectilineal figure is said to be circuscribed Fig. 1, about a circle, when each of its sides is a tangent to the circle. 3, A circle is said to be inscribed in a rectilineal Fig. 1. figure, when each side of the figure is a tangent to the circle. 4. A circle is said to be circumscribed about a recti- Fig. 1. lineal figure, when the circumference passes through the vertex of each angle of the figure. 5. A right line is said to be inscribed in a circle, Fig. 1. when its extremities are in the circumference of the circle. ta Fig. 2. In a given circle (BCA) to inscribe a right line equal to a given right line (D), which is not greater than the diameter of the circle. Draw a diameter AB of the circle, and if this is equal to the given line D, the problem is solved. ( 1 )Prop.3. If not, take in it the segment AE equal to D (1), from the centre A with the radius AE describe a (2)Def. 13. circle ECF, and draw to either intersection of it 1 3 B.). LE (1) 2) eqi EL for to (4) san AL ang tria is e PROP. II. PROB, BD Itra Fig. 3. In a given circle (BAC) to inscribe a triangle equi angular to a given triangle (EDF). Draw the line GH a tangent to the given circle in any point A at the point A with the line AH make the angle HAC equal to the angle E, and at the same point with the line AG make the angle GAB equal to the angle F, and draw BC. (1) Consir. Because the angle E is equal to HAC (1), and HAC is equal to the angle B in the alternate seg(2) Prop.32 ment (2), the angles E and B are equal, also the angles F and C are equal, therefore the remaining (3) Cor. 2. angle D is equal to BAC (3), and therefore the triProp.32. angle BAC inscribed in the given circle is equiangular B. 3. DB the DE B. I. i bei ther PROP. III. PROB. 3) Fig. 4. Ea About a given circle. (ABC) to circumscribe a triangle equiangular to a given triangle (EDF). Produce any side DF of the given triangle both ways to G and H, from the centre K of the given circle the B, 3. B, 1. draw any right line KA, with this line at the point K make the angle BKA equal to the angle EDG, and at the other side of KA make the angle AKC equal to EFH, and draw the lines LM, LN, and MN tangents to the circle in the points B, A, and C. Because the four angles of the quadrilateral figure LBKA taken together are equal to four right angles (1), and the angles KBL and KAL are right angles (1) Cor. 6. (2), the remaining angles AKB and ALB are together P.32. B. I. equal to two right angles, but the angles EDG and (2) Prop.18 EDF are together equal to two right angles (3), there- (3) Prop-13 fore the angles AKB and ALB are together equal B. 1. to EDG and EDF, but AKB and EDG are equal (4) Constr. (4), and therefore ALB and EDF are equal. In the same manner it can be demonstrated that the angles ANC and EFD are equal, therefore the remaining angle M is equal to the angle E (5), and therefore the (5) Cor. 2. triangle LMN circumscribed about the given circle Prop. 32. is equiangular to the given triangle. PROP. IV. PROB. In a given triangle (BAC) to inscribe a circle. Fig. 5. Bisect any two angles B and C by the right lines BD and CD, and from their point of concourse D draw DF perpendicular to any side BC, the circle described from the centre D with the radius DF is inscribed in the given triangle. Draw DE and DG perpendicular to BA and AC. In the triangles DEB, DFB the angles DEB and DBE are equal to the angles DFB and DBF(1), and (1) Const. the side DB is common to both, therefore the sides DE and DF are equal (2): in the same manner it can (2) Prop.26 be demonstrated that the lines DG and DF are equal, therefore the three lines DE, DF and DG are equal (3), and therefore the circle described from the cen- (3) At. 1. tre D with the radius DF passes through the points E and G, and because the angles at F, E and Gare right, the lines BC, BA and AC are tangents to the circle (4), therefore the circle FEG is inscribed in (4) Prop. 16 the given triangle. •B. 3. See N. B. I. |