PROPOSITION XXXIII. PROBLEM.-On a given finite straight line (AB) to describe a segment of a circle which shall contain an angle equal to a given rectilineal angle (C). Let the given angle be a right angle: SOLUTION. Bisect AB in F (a), and from the center F, with the radius AF, describe the semicircle AHB, and it will be the segment required. DEMONSTRATION. For the angle H in the semicircle AHB is a right C F I. 10. III. 31. H angle (6), and the given angle is a right angle; therefore the angle in the segment AHB is equal to the given angle α. If the given angle be not a right angle: SOLUTION. At the point A in the straight line AB form the angle BAD equal to the given angle C (c), and from the point A draw AE perpendicular to AD (d); bisect AB in F (a), and from F draw FG perpendicular to AB (d); then from the center G, with the radius GA, describe the circle ABEH, and join GB and BE. AHB shall be upon the given line AB, and shall equal to the given angle C. (c) I. 23. Then the segment contain an angle DEMONSTRATION. Because in the triangles AFG and BFG the side AF is equal to BF, the side FG common to both, and the angle AFG equal to BFG, the base AG is equal to GB (e); therefore the circle AHE, described from G as a center, with the radius GA, shall pass through the point B; and because from the point A, at the extremity of the diameter AE, AD is drawn perpendicular to AE, therefore AD touches the circle (f); and because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (g); but the angle DAB is equal to the angle C; therefore the angle in the segment AHB is equal to the angle C, and it is upon the straight line AB PROPOSITION XXXIV. PROBLEM.-To cut off from a given circle (ABC) a segment which shall contain an angle equal to a given rectilineal angle (D). SOLUTION. Draw the straight line EF, touching the circle ABC in the point B (a), and at the point B in the straight line EF form the angle FBC equal to the given angle D (b); then the segment BAC cut off of the given circle shall contain an angle equal to the given angle D. (a) III. 17. DEMONSTRATION. Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal to the angle in the alternate segment BAC (c); but the angle FBC is equal to the angle D; therefore the angle in the segment BAC cut off from the given circle is equal to the given angle D. PROPOSITION XXXV. THEOREM.-If two straight lines (AC and BD) cut one another within a circle, the rectangle under the segments (AE and EC) of one of them is equal in area to the rectangle under the segments (BE and ED) of the other. DEMONSTRATION. [1.] Let the two lines intersect in the center of the circle; then since AE, EC, BE, and ED are all equal (a), the rectangle under AE and EC must be equal to the rectangle under BE and ED. B A (a) Def. 15 [2] Let one of the given lines AC pass through the center, and the other not; find the center F of the circle (b), and join DF and BF. Then in the triangle FBD the rectangle under BE and ED is equal in area to the difference of the squares on BF and FE (c), that is, to the difference of the squares on CF and FE; then because AE is equal to the sum of the two lines CF and FE, and CE is equal to their difference, the rectangle under CE and AE is equal in area to the difference of the squares on CF and FE (d); therefore the rectangle under BE and ED is equal in area to the rectangle under CE and AE. [3.] Let neither of the given lines pass through the center; find the center F (b), and through E and F draw the diameter HG; then the rectangle under HE and EG has just been shown to be equal in area to the rectangle under AE and EC, and also equal in area to the rectangle under BE and ED; therefore the rectangle under AE and EC is equal in area to the rectangle under BE and ED. SCHOLIUM. By a modification in the principle of demonstrating the second case, it is made to include the second and third cases of Simson, and thus reduces the whole number of cases from four to three, and greatly abridges the proof. PROPOSITION XXXVI. THEOREM.-If from a point (D) without a circle two straight lines be drawn, one of which (DA) cuts the circle, and the other (BD) touches it, the rectangle under the whole line which cuts the circle (DA) and the segment without the circle (DC), is equal in area to the square on the line (BD) which touches it. DEMONSTRATION. [1.] Let the line DA pass through the center E of the circle, and join EB; then in the right-angled triangle DBE the square on BD is equal in area to the difference of the squares on DE and BE (a), or to the difference of the squares on DE and CE; but because DA is equal to the sum of the two lines (b) II. 5, cor. 1. DE and CE, and DC is equal to their difference, the rectangle under DA and DC is equal in area to the difference of the squares on DE and CE (6); therefore the square on BD is equal in area to the rectangle under DA and DC. [2.] If the line DA does not pass through the center, find the center E (c), and join EA, EB, EC, and ED. Then in the isosceles triangle ACE the rectangle under DA and DC is equal in area to the difference of the squares on DE and CE (d), or to the dif ference of the squares on DE and EB; but in the right-angled triangle DEB the square on DB is equal in area to the difference of the squares on DE and EB; therefore the square on DB is equal in area to the rectangle under DA and DC. COROLLARY 1. THEOREM. If from a point (A) without a circle two straight lines (AB and AC) be drawn cutting it, the rectangles under the whole lines and the parts of them without the circle are equal in area to one another (AE and AB to AF and AC). For each of them is equal in area to the straight line AD drawn from the same point to touch the circle (a), and therefore they are equal in area to one another D III. 1. (d) II. 6, cor. 2. B (a) III. 36. COROLLARY 2. THEOREM. If the rectangles under the segments (AE, EC, and BE, ED), made by the intersection of the diagonals of a quadrilateral figure (ABCD), are equal in area, or if the rectangles under the segments (EA, ED, and EB, EC), made by producing its opposite sides to intersect, are equal in area, the quadrilateral may have a circle described about it. F B Describe a circle passing through three of the angles A, B, and C of the quadrilateral (a), then (a) Theor. attached to III. 1. B (b) III. 35, or 36, cor. Hypoth. it shall also pass through the fourth angle D; for if not, let it cut ED in F, then the rectangle under AE and EC is equal in area to the rectangle under BE and EF (b); but the rectangle under AE and EC is also equal in area to the rectangle under BE and ED (c); therefore the rectangle under BE and EF is equal in area to the rectangle under BE and ED, the less equal to the greater, which is absurd; therefore the circle passes through the four angles of the quadrilateral figure ABCD. PROPOSITION XXXVII. THEOREM.-If from a point (D) without a circle two straight lines be drawn, one (DA) cutting the circle, and the other (DB) meeting it, and if the rectangle under the whole line which cuts the circle, and the part of it without the circle (DA and DC), be equal in area to the square on the line (DB) which meets it, that line touches the circle. CONSTRUCTION. Draw the straight line DE, touching the circle (a), find the center F of the circle (b), and join FB, FD, and FE. DEMONSTRATION. Then the rectangle under DA and DC is equal in area to the square on DE (c); but the rectangle under DA and DC is also equal in area to the square on DB (d); therefore the square on DB is equal to the square on DE, and DB is equal to DE (e); then in the triangles FBD and FED, because the side DB is equal to DE, the side BF to FE, and the side DF common to both, the angle B is equal to the angle E (ƒ); but the angle E is a right angle (9), therefore the angle B is a right angle and the straight line DB meets the circle (h) (a) III. 17. I. 46, cor. 2. Construct III. 16. E |