Similarly it may be shown that AD. BC=square on the transverse common tangent. EXERCISES. 1. The greater of two chords in a circle is nearer the centre than the other. 2. Of two chords unequally distant from the centre the one nearer the centre is the greater. 3. AB is the diameter of a circle, and P, Q any two points on the curve. AP and BQ intersect in C, and AQ and BP Then in C'. AP.AC+BQ.BC=AC'. AQ+BC'.BP. 4. Two chords of a circle, AB and CD, intersect in O and are perpendicular to one another. If R denotes the radius of the circle and E its centre, 8R2=AB2+CD2 + 40E2. 5. Circles are described on the four sides of a quadrangle as diameters. The common chord of any two adjacent circles is parallel to the common chord of the other two. 6. A circle S and a line L, without one another, are touched by a variable circle Z. The chord of contact of Z passes through that point of S which is farthest distant from L. 7. ABC is an equilateral triangle and P is any point on its circumcircle. Then PA + PB+PC=0, if we consider the line crossing the triangle as being negative. 8. CD is a chord parallel to the diameter AB, and P is any point in that diameter. Then PC2+PD2= PA2+ PB2. CONSTRUCTIVE GEOMETRY. 180°. Problem.-AB being a given segment, to construct the segment AB/2. Also SECTION V. Constr. -Draw BCL to AB and equal to it. Proof. Since ABC is right-angled at B, Cor. The square on the diagonal of a given square is equal to twice the given square. 181°. Problem.-To construct AB√3. Constr.-Take BC in line with AB and equal to it, and on AC construct an equilateral triangle ADC. (124°, Cor. 1) BD is the segment AB√3. Proof.-ABD is a ̃ ̄ ̄|, and AD=AC=2AB. BD2=3AB2, and BD=AB√3. Cor. Since BD is the altitude of an equilateral triangle and AB is one-half the side, .. the square on the altitude of an equilateral triangle is equal to three times the square on the half side. But Proof. Since LB is a right angle, AC2=AB2+BC2. and 182°. Problem.-To construct AB√5. Constr. -Draw BCL to AB and equal to twice AB. Then AC is the segment AB/5. A B BC2=4AB2; AC2=5AB2, AC=AB√5. B C (169°, Cor. 1) B 183°. The three foregoing problems furnish elements of construction which are often convenient. A few examples are given. A Ex. 1. AB being a given segment, to find a point C in its line such that AC2=AB. CB. B Analysis AC2=AB. CB=AB(AB – AC), AC2+AC. AB=AB2. Considering this an algebraic form and solving as a quadratic in AC, we have AC (AB√5- AB), and this is to be constructed. Constr.-Construct AD=AB√5 (by 182°) as in the figure, C B and let. E be the middle point of BD. Take DF=DE. Then AF AB/5-AB; .. bisecting AF in G, AG=AC A E =(AB√5-AB), and the point C is found. Again, since 5 has two signs or F' take its negative sign and we have AC (AB/5+AB). Therefore, for the point C', on AD produced take DF'=DE, and bisect AF' in G'. Then AG'=(AB√5+AB); and and since AC' is negative we set off AG' from A to C', C' is a second point. The points C and C' satisfy the conditions, AC2=AB. CB and AC"=AB. C'B. A construction effected in this way requires no proof other than the equation which it represents. It is readily proved however. For It will be noticed that the constructions for finding the two points differ only by some of the segments being taken in different senses. Thus, for C, DE is taken from DA, and for C', added to DA; and for C, AC is taken in a positive sense equal to AG, and for C', AC' is taken in a negative sense equal to AG'. In connection with the present example we remark :— I. Where the analysis of a problem involves the solution of a quadratic equation, the problem has two solutions corresponding to the roots of the equation. 2. Both of the solutions may be applicable to the wording of the problem or only one may be. 3. The cause of the inapplicability of one of the solutions is commonly due to the fact that a mathematical symbol is more general in its significance than the words of a spoken language. 4. Both solutions may usually be made applicable by some change in the wording of the problem so as to generalize it. The preceding problem may be stated as follows, but whether both solutions apply to it, or only one, will depend upon our definition of the word "part." See Art. 23°. To divide a given segment so that the square upon one of the parts is equal to the rectangle on the whole segment and the other part. Def.--A segment thus divided is said to be divided into extreme and mean ratio, or in median section. Ex. 2. To describe a square when the sum of its side and diagonal is given. Analysis. If AB is the side of a square, AB√2 is its diagonal, (180°) .. AB(1 +√2) is a given segment =S, say. Then AB=S(/2-1). E S F Constr.-Let EF be the given segment S. Draw FG and to EF, and with centre G and radius GF describe a cutting EG in H and H'. H' EH is the side of the square; whence the square is easily constructed. If we enquire what EH' means, we find it to be the side of the square in which the difference between the side and diagonal is the given segment S. The double solution here is very suggestive, but we leave its discussion to the reader. A 184°. Problem.-To find a segment such that the rectangle on it and a given segment shall be equal to a given rectangle. P G E D .. = On DA produced make AP=S, and Proof. Complete the □s PEQD, PGBA, and BCQF. Constr.--Let S be the given segment, and AC the given rectangle. Def. The segments AP and CQ are reciprocals of one another with respect to the AC as unit. B 185°. Problem.-To find the side of a square which is equal to a given rectangle. Constr.-Let AC be the rectangle. Make BE=BC and in line with BA. JE On AE describe a semicircle, and produce CB to meet it in F. BF is the side of the required square. Proof.- Since AE is a diameter and FB a half chord to it, |