HINT. Circumscribe circles, produce bisectors until they meet the cir cumference, and join the point of intersection to one end of the base. (Exs. 1061 and 1383.) Ex. 1385. If two angle bisectors of a triangle are equal, the triangle is isosceles. (Ex. 1384.) * Ex. 1386. The square of the side of a regular pentagon increased by the square of one of its diagonals is equal to five times the square of the radius. (Ex. 1382.) * Ex. 1387 A regular pentagon is equivalent to a rectangle having one side equal to five times the radius, and the other to of a diagonal of the pentagon. * Ex. 1388. The product of the diagonals of an inscribed quadrilateral is equal to the sum of the products of the opposite sides. Theorem.) (Ptolemy's Ex. 1389. An equiangular polygon inscribed in a circle is regular if the number of sides is odd. Ex. 1390. If circles be circumscribed about the four triangles into which a quadrilateral is divided by its diagonals, their centers form the vertices of a parallelogram. * Ex. 1391. Each angle formed by joining the feet of the three altitudes of a triangle is bisected by the corresponding altitude. (Ex. 581.) * Ex. 1392. If from any point in the circumference of a circle perpendiculars be dropped upon the sides of an inscribed triangle (produced, if necessary), the feet of the perpendiculars are in a straight line. (Ex. 581.) Ex. 1393. To construct a triangle having given the feet of its three altitudes. * Ex. 1394. If Pn and P are respectively the perimeters of an inscribed and circumscribed regular polygon of n sides, and p2n, and P2n the perimeters of regular polygons of 2 n sides respectively inscribed and circumscribed about the same circle, prove that P2 is the harmonical mean between Pn and Pn; i.e. P21 = 2 P2 P2 * Ex. 1395. Using the notations of Ex. 1394, prove that pan is the mean proportional between p1 and P2; i.e. p2n = √PnP2n APPENDIX TO PLANE GEOMETRY ALGEBRAIC SOLUTIONS OF GEOMETRICAL PROPOSITIONS. MAXIMA AND MINIMA PRACTICAL PROBLEMS CONSTRUCTION OF ALGEBRAIC EXPRESSIONS 442. NOTE. In the following propositions, a, b, c, d, etc., denote given lines, while x, y, z, etc., denote required lines. HINT. x is the mean proportional between a and b (325). Ex. 1396. Construct x = 2√3 by means of an equilateral triangle. 444. Complex algebraic expressions are constructed by means of the eleven constructions of (443). It is quite often necessary to transform the algebraic expressions, in order to make them special cases of (443). Different algebraic transformations lead to different solutions. or i.e. x is the mean proportional between 3 a and b, x = =√ab√3, i.e. find Vab by means of (443, 7), and √ab√3 by means of (Ex. 1396). i.e. find the mean proportional between b and c, and construct a right triangle, having one arm equal to the mean proportional, and the hypotenuse equal to a. 445. REMARK. The expressions to be constructed in Geometry are always homogeneous, and either of the first or second degree. ab Some expressions may be made homogeneous by the introduction of unity. Thus x = ab, may be written x = i.e. x is the fourth proportional to 1, a, and b. " 1 446. The impossibility of the construction is indicated if: (1) The expression is imaginary. (2) The value of x is not within the limits indicated by the problem. (3) Sometimes, if the result is negative. If the required line has a certain direction, a negative result would indicate a line of opposite direction. Ex. 1401. Construct the following expressions : (a) √2 ab, (b) a√5, (c) a + 3b, (d) √4a2 = b2. SOLUTION OF PROBLEMS BY MEANS OF ALGEBRAIC 447. If a problem requires the construction of lines, it is often possible to state the condition in the form of an equation. The solution of the equation gives the unknown line in an algebraic form, which may be constructed according to (444). Ex. 1402. To divide a line externally so that the small external segment is the mean proportional between the other segment and the given line. Analysis. Let x be one segment, then a +x is the other, and a + x:x=x: α. As the minus sign would give a negative answer, i.e. an internal segment, we have to construct Draw ABα. At B, draw BC perpendicular to AB and equal to 2. a Draw AC, and produce it to D, so that CD= CB. Then x= AD. .. produce BA to E, so that AE = AD. E is the required point. 448. The analysis used in this problem is called algebraic analysis in distinction from the purely geometric analysis given in Book I. The algebraic analysis contains a proof for the correctness of the construction, although quite often a purely geometric proof may be found. |