PROP. XVII. PROB. To divide a straight line, whether internally or externally, so that the rectangle under its segments shall be equivalent to a given rectangle. Let AB be the straight line which it is required to cut, so that the rectangle under its segments shall be equivalent to a given rectangle. From the extremities of AB, erect the perpendiculars AD and BE, equal to the sides of the given rectangle, and in the same or in opposite directions, according as the line is to be cut intermally or externally; join DE, on which, as a dia- ID meter, describe a circle, meeting AB or its exten- | sion in the point C: AC and CB are the segments N _><2. required. =ö7 For join DC and CE. The angle DCE, being contained in a semicircle, is a right angle (III. 19.), and therefore, in both cases, the angles ACD and BCE are together equal to a right angle. But the angles ACD and CDA are likewise together equal to a right angle (I. 30. cor. 1.); and consequently the angles BCE and CDA are equal. Wherefore the right-angled triangles CBE and CAD, having the acute angle ADC equal to BCE, are similar (VI. 11.); whence AC : AD : : BE : CB, and (V. 6.) AC.CB=AD.B.E. Scholium. It is obvious that, in the second case, the circle, lying on both sides of the given line AB, must always intersect its extension in two points C and C. But, in the first case, the circle may either cut AB in two points C and C, or touch it in a single point, which will hence mark a limitation of the problem. A straight line drawn from the centre of the circle parallel to AD or BE, must (VI.1.). divide AB proportionally, and hence bisect it; but that parallel would also be perpendicular (I. 22.) to AB, and therefore (III.4.) bisect the chord CC. Consequently the points C and C are equally distant from the middle of AB, and the portion AC is equal to BC’. When these points come to coincide, they must therefore pass into the middle point of AB, or that of its contact with the circle. When the circle does not reach AB, the problem fails, because (II. 17. cor. 1.) no straight line can be divided internally, such that the rectangle under the segments shall exceed the square of its half. This impossibility is indicated by the circle not reaching the straight line AB. This proposition furnishes one of the simplest and most elegant methods for constructing quadratic equations; the segments of the line denoting the roots, and indicating by position their character. The first case has two additive, roots, which may become equal or merge in a single root, then limiting the possibility of the equation; the second case has always two unequal roots, the one additive and the other subtractive. In both cases, those roots, conjoined in their actual position, complete the line AB. N z PROP. XVIII. THEOR. The rectangle under any two sides of a triangle is equivalent to the rectangle under the perpendicular let fall on the base and the diameter of the circumscribing circle. Let ABC be a triangle, about which is described a circle having the diameter BE; the rectangle under the sides AB and BC is equivalent to the rectangle under BE and the perpendicular BD let fall from the vertex of the triangle upon the base AC. For join CE. The angle BAD is equal to BEC (III. 16.), since they both stand upon the same arc BC; and the angle ADB, being a right angle, is (III. 19.) equal to ECB, which is contained in a semicircle. Wherefore the triangles ABD and EBC, being thus similar (VI. 11.), AB : BD: ; EB : BC, and consequently (V. 6.) AB.BC=EB.BD. PROP. XIX. THEOR. The square of a straight line that bisects, whether internally or externally, the vertical angle of a triangle, is equivalent to the difference between the rectangle under the sides, and the rectangle under the segments into which it divides the base. In the triangle ABC, let BE bisect the vertical angle CBA or its adjacent angle CBF ; then BE*=AB.BC— AE.EC, or AE.EC–AB.BC. For (III. 9. cor.) about the triangle describe a circle, produce BE to the circumference, and join CD. Theangles BAE and BDC, standing upon the same arc BC, are (III. 16.) equal, and the angle ABE is, by hypothesis, equal to DBC; wherefore (VI. 11.) the triangles AEB and DCB are similar, and AB : BE : : DB : BC. Consequently (V. 6.) AB.BC= BE.BD ; but BE.BD = BE.ED + BE", or BE.ED–BE", and (III. 26.) BE.ED=AE.EC; wherefore AB.BC=AE.EC+BE*, or AE.EC–BE*; and consequently BE*=AB.BC—AE.EC, or AE.EC— AB,BC. PROP. XX. THEOR. The rectangles under the opposite sides of a quadrilateral figure inscribed in a circle, are together equivalent to the rectangle under its diago. mals. In the circle ABCD, let a quadrilateral figure be inscribed, and join the diagonals AC, BD; the rectangles AB, CD and BC, AD, are together equivalent to the rectangle AC, BD. o z Tor (I. 4.) draw BE, making an angle ABE equal to CBD. The triangles AEB and DCB, having thus the angle ABE equal to DBC, and the angle BAE or BAC equal (III. 16.) to BDC, are similar (VI. 11.), and hence AB : AE :: BD: CD; whence (V.6.)AB.CD=AE.BD. Again, because the angle ABE is equal to DBC, add EBD to each, and the whole angle ABD is equal to EBC; and the angle ADB is equal to ECB (III. 16.); wherefore the triangles DAB and CEB are similar (VI. 11.), and AD : BD :: EC : BC, and consequently BC.AD=EC.BD. Whence the rectangles AB, CD and BC, AD are together equal to the rectangles AE, BD and EC, BD, that is, to the whole rectangle AC, BD. PROP. XXI. THEOR. Triangles which have a common angle, are to each other in the compound ratio of the containing sides. Let ABC and DBE be two triangles, having the same or an equal angle at B ; ABC is to DBE in the ratio compounded of that of BA to BD, and of BC to BE. For join AE and CD. The ratio of the triangle ABC to DBE may be conceived as compounded of that of ABC to |