To cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle. Given the circle ABC, and the rectilineal angle D; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D. (Const.) Draw the straight line EF touching the circle ABC in the point B (III. 17), and at the point B, in the straight line BF, make the angle FBC equal to drawn from the point of contact B, is equal to the angle D; the angle in the to the angle D; segment BAC is equal E B F wherefore the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. If two chords in a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Given the two chords AC and BD, within the circle ABCD, cutting one another in the point E; to prove that the rectis equal angle contained by AE and EC, to the rectangle contained by BE and ED. Case First.-If AC and BD pass each of them through the centre, so that E is the centre; it is evident that and ED, being all equal, A E D AE, EC, BE, B the rectangle AE EC, is likewise equal to the rectangle BE ED. Case Second.-But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles, in the point E; then if BD be bisected in F, F is the centre of the circle ABCD; join AF; (Dem.) and because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E; AE and EC are equal to one another (III. 3); and because the straight line BD is cut into and into two unequal in the two equal parts in the point F, point E, that D the rectangle BE ED, together with the square on EF, is equal to the square on FB (II. 5); is, to the square of FA; but the squares on AE and EF are equal to the square on FA (I. 47); therefore the rectangle BE ED, together with the square on EF, is equal to the squares on AE and EF. Take away the common square on EF, and the remaining rectangle BE ED is equal to the remaining square on AE; that is, to the rectangle AE EC, since AE is equal to EC. A E B Q F and from F draw FG Join AF, therefore AG (Dem.) wherefore D F E A G B Case Third.-Let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles; then, as before, if BD be bisected in F, is the centre of the circle. perpendicular to AC (I. 12); is equal to GC (III. 3); the rectangle AE EC, together with the square on EG, is equal to the square on AG (II. 5). To each of these equals add the square on GF; therefore the rectangle AE EC, together with the squares on EG and GF, is equal to the squares on AG and GF. But the squares on EG and GF are equal to the square on EF; and the squares on AG and GF are equal to the square on AF; therefore the rectangle AE EC, together with the square on EF, is equal to the square on AF; that is, to the square on FB. But the square on FB is equal to the rectangle BEED, together with the square on EF (II. 5); therefore the rectangle AE EC, together with the square on EF, is equal to the rectangle BEED, together with the square on EF. Take away the common square on and the remaining rectangle AE⚫EC EF, H F Case Fourth.-Let neither of the straight lines AC, BD pass through the centre; take the centre F, and through E, the intersection of the straight lines AC and DB, draw the diameter GEFH; (Dem.) and because the rectangle GE EH is equal, by Case Third, to the rectangle AE· EC; rectangle BEED; therefore the rectangle AE EC is equal to the rectangle BE⚫ ED. and also to the ** See Appendix-Proposition (D). If from a point without a circle a secant and a tangent to it be drawn, the rectangle under the secant and its external segment is equal to the square on the tangent. and DCA and of which DCA cuts the the rectangle Given any point D without the circle ABC, First, (Dem.) there Either DCA passes through the centre, or it does not. let it pass through the centre E, and join EB; fore the angle EBD is a right angle (III. 18); and because the straight line AC is bisected in E, and produced to the point D, the rectangle AD DC, together with the square on EC, is equal to the square on ED (II. 6); and CE is equal to EB; therefore the rectangle AD DC, together with the square on EB, is equal to the square on ED; B D E A but (I. 47) the square on ED is equal to the squares on EB and BD, because EBD is a right angle; therefore the rectangle AD DC, together with the square on EB, is equal to the squares on EB and BD; take away the common square on EB; therefore the remaining rectangle AD DC is equal to the square on the tangent DB. Second, when DCA does not pass through the centre of the circle ABC, take the centre E (I. 3), and draw EF perpendicular to AC (I. 12), and join EB, EC, and ED; (Dem.) and because the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it shall likewise bisect it (III. 3); therefore AF is equal to FC; therefore the rectangle AD DC, together B F with the squares on CF and FE, is equal to the squares on DF and FE; but the square on ED is equal to the squares on DF and FE, because EFD is a right angle; and the square on EC is equal to the squares on CF and FE; rectangle AD DC, together with the square on EC, to the square on ED; and CE is equal to EB; therefore the is equal therefore the rectangle AD DC, together with the square on EB, is equal to the square on ED; take but the squares on EB and BD are also equal to the square on ED, because EBD is a right angle; therefore the rectangle AD DC, together with the square on EB, is equal to the squares on EB and BD; away the common square on EB; fore the remaining rectangle AD · DC is equal to the square on DB. there COR.-If from any point without a circle there be drawn two straight lines cutting it, as AB and AC, the rectangles contained by the whole lines and the parts of them without the circle are equal to one another-namely, the rectangle BA AE, to the rectangle CAAF; (Dem.) for each of them is equal to the D E A square of the straight line AD which touches the circle. Schol. 1.-The above corollary and proposition 35, may be enunciated thus: The rectangles under the segments of two intersecting chords of a circle are equal, whether they cut internally or externally. Schol. 2.—The first case of this proposition affords a geometrical principle by which the length of the diameter of the earth may be computed. 1 Thus, if CD (first fig.) represent the height of a mountain, which, of course, is greatly exaggerated in size compared with AC, which represents the diameter of the earth; and if this height is known, and also the distance DB of the horizon, then since AD DC = DB2, it follows that the product of the numbers representing the lengths of CD and AD is equal to the square of the number denoting the length of DB; and this square number being divided by CD, gives AD for a quotient, and then AC is known. This method, however, does not afford very precise results (see PRACTICAL MATHEMATICS). The proposition is also the geometrical principle on which the method of levelling is founded. PROPOSITION XXXVII. THEOREM. If from a point without a circle a secant be drawn, and also a line meeting the circle, and if the rectangle under the secant and its external segment be equal to the square on the other line, this line will be a tangent. Given any point D without the circle lines DCA and DB drawn from it, circle, and DB meets it; to prove that ABC, and two straight of which DCA cuts the if the rectangle AD⚫ DC be equal to the square on DB, DB touches the circle. (Const.) Draw the straight line DE, touching the circle ABC (III. 17); FD; DE (III. 36); D and the find the centre F, and join FE, FB, and (Dem.) then FED is a right angle (III. 18); because DE touches the circle ABC, and DCA cuts it, rectangle AD DC is equal to the square on but the rectangle AD⚫ DC is equal to the given square on DB; therefore the square on DE is equal to the square on ᎠᏴ ; and the straight line DE is equal to the straight line DB; and FE is equal to FB, wherefore DE and EF are equal to DB and BF; and the base FD is common to the two triangles DEF and DBF; therefore the angle DEF is equal to the angle DBF (1.8); and DEF is a right angle; therefore also DBF is a right angle; but FB is B A F a radius, and the straight line which is drawn at right angles to a radius, from the extremity of it, touches the circle (III. 16, Cor.); therefore DB touches the circle ABC. COR.-Hence two tangents drawn to a circle from the same point are equal. (Dem.) For DE and DB are two tangents, and the square on each is equal to the rectangle AD DC; therefore DE2 = DB2, and consequently DE = DB. If from the extremities of the base of a triangle perpendiculars be drawn to the opposite sides, the line joining their intersection and the vertex is perpendicular to the base. Given a triangle ABC, and AF, BE perpendiculars upon the sides, intersecting in G; to prove that CGD is perpendicular to the base AB. 3. (Const.) Join EF. (Dem.) Since the quadrilateral figure CEGF has the angles at E and F right angles, a circle may be described about it (III. 22), therefore the angle GEF is equal to the angle GCF (III. 21); again, since the angles A AEB and AFB are right angles, a circle described on AB as diameter would pass through E and F (III. 31), therefore the angle BEF, that is, GEF is equal to the angle BAF (III. 21), wherefore the angles GAD and GCF are equal, each being equal to GEF; but the angles AGD and CGF are also equal (I. 15), therefore the triangles AGD and CGF have two angles of the one equal to two angles of the other, hence the angle ADG is equal to the angle CFG, D but |