be drawn, the tangent will óe a mean proportional between the entire secant and the part without the circle--, Thus if O F 48 (fig. 48) be the point, O A the tangent, and 0 C the secant (40), then we say that 0 A is a mean proportional between 0 D and 0 C, that is O D:0 A::0 A:0 C. For draw the chord A D, and the triangles O A D and o A C will be similar. Why?-_Because they have the angle 0 common, and O A D=A CD, since both have for their measure half the arc A D (41). Then 0 D is homologous to 0 A, and 0 A is homologous to 0 C, and we have O D : 0 A::0 A:0 C. 82. – To divide a given line in extreme and mean ratio—. By this we mean to divide a line into two such parts that the greater part shall be a mean proportional beF 49 tween the whole and the less. Thus A B (fig. 49) will be divided in extreme and mean ratio, if we can find a point F such that B F:AF::AF:A B. The question then is to find the point F. Erect the perpendicular B C=half of A B, and with C as a centre and C B as radius describe a circle. Through A and C draw the secant A E. Then with A as a centre and A D as a radius describe an arc cutting A B in F. F will be the point sought. For by the preceding proposition we have A D:AB::A B:A E. Then (68) A B-A D:A E-A B :: A D:A B. But A B-A D=B F, and A E-A B=A E-D E=A D=AF. Therefore BF:AF:: A F: A B, which is the proportion sought. Quadrilaterals. 83. Any figure bounded by four straight lines is called a quadrilateral. Quadrilaterals take different names ac cording to the relations of their sides and angles. If the F 50 opposite sides are parallel (fig. 50) the quadrilateral is a F 51 parallelogram. If only two of the sides are parallel, (fig. 51) the quadrilateral is a trapezoid. If no two of the sides are F 52 parallel, (fig. 52) the quadrilateral is a trapezium. Of these three classes, the parallelogram is most important. Parallelograms are either right or oblique. Right parallelograms have all their angles right angles. Of these there are two denominations, the square, and the oblong or rectanF53 gle. The square has all its sides equal (fig. 53). The F 54 oblong has only its opposite sides equal (fig. 54.) Oblique parallelograms have none of their angles right angles. Of these there are also two denominations, the rhombus, and the rhomboid. The rhombus has all its sides equal (fig. 55). F55 The rhomboid has only its opposite sides equal (fig. 56). F 56 It will be observed that the fundamental property of a parallelogram, namely, the parallelism of the opposite sides, is common to the four last mentioned figures. There is also another property resulting from this (38), namely, the equality of the opposite sides. When we use the word parallelogram alone, we shall include in it only these two properties. 84. — The diagonal of a parallelogram divides it into two equal triangles By diagonal we mean a straight line joining two vertices not adjacent. Thus B D (fig. 56) is F 56 a diagonal. Now we say that the triangles B A D and B C D are equal. Why?--Because A B=D C and A D=B C, by the definition, and B D is common to both. Therefore (57) they are equal. Moreover-The opposite angles of a parallelogram are equal— Why?-Because A=C, beiug opposite to equal "sides in equal triangles; and since A B D=B D C (34) and A D B D B C, we have A BD+D BC=A D B+B DC, or A B C=A D C. Again-The sum of the angles of a parallelogram is equal to four right angles— Why?-Because (46) the sum of the angles in each of the triangles is equal to two right angles. Lastly— The two diagonals of a parallelogram mutually bisecl each other Why?-Because the triangles A B E and C D E (fig. 55) are equal (55), since A B=C F 55 D, the angle A B E=E D C (34), and B A E=E C D. Therefore A E=E C, and B E=E D. Polygons. 85. Polygon is the general name for every figure bounded by straight lines. Accordingly we might have treated of triangles and quadrilaterals under this head. But we thought it more useful as well as more simple, to consider them separately. Polygons are divided into regular and irregular. Regular polygons are those which have all their sides equal, and all their angles equal. Thus an equilateral triangle and a square are regular polygons. Irregular polygons are such as do not possess both these properties. Similar polygons are those which have their angles equal, each to each, and their homologous sides proportional. The student must observe that two polygons may be equiangular with respect to each other, when peither is equiangular considered by itself. Thus a regular polygon is equiangular in itself, but two similar polygons are not necessarily so, though they are equiangular with respect to each other. There are particular names for polygons depending upon the number of sides. Thus a pentagon is a polygon of five sides, a hexagon one of six sides, a decagon one of ten sides, &c. But we shall use the general term polygon, unless where the necessity of the case requires us to be more specific. 86. The sum of the interior angles of any polygon is equal to as many times two right angles as there are sides minus F57 two-- Why ?-Because if from any vertex as A (fig. 57) diagonals be drawn to all the vertices not adjacent, the polygon will be divided into as many triangles as there are sides minus two. Thus if the polygon have six sides, there will be four triangles, and so of any other number. Let it be observed that we here speak only of convex polygons, that is of those whose vertices are all directed outward, as in the figure. 87. --If two polygons are composed of the same number of similar triangles similarly disposed, the polygons are simF 58 ilar. Thus if the two polygons (fig. 58) are composed of the same number of similar triangles, then we say they have their angles equal, each to each, and their homologous sides proportional. Why? Because if A B C is similar to F G H, then the angle B=G, and A B:F G::BC:G H (78). Again the angle B C D=G H I, because from the similarity of the successive triangles, B Ź CA=G HF and A CD=F HI; wbile from the two proportions B C:GH::AC:FH and A C:FH::C DOH I, we have (64) BC:GH::CD:HI. Thus far then, we have the angles equal, each to each, and the homologous sides proportional, and it is evident that the same reasoning might be continued as long as there were similar triangles placed in the same order. 88. -Upon a given line to construct a polygon similar to a F 58 given polygon Suppose F G (fig. 58) were the given lipe and A B C D E the given polygon. Consider FG homologous, for example, to A B. Then draw G H making the angle F G HÓA B C, and dra :v FH making the angle G F H=B A C. The triangles A B C and F G H will be similar. Again draw H I making the angle FH I=A C D, and draw F I making the angle H FI=C A D. The triangles A CD and FHI will be similar. Proceed in this aiander till the construction is completed, and the two polygons will be similar by the preceding proposition. 89. -Two regular polygons of the same number of sides are similar- Suppose the two polygons are regular hexagons (fig 57). Then we say, in the first place, they are F57 equiangular with respect to each other, for each of the angles in both polygons is equal to one sixth of eight right angles (86). Again their homologous sides are proportional. For, by the definition of regular polygons (85) A B=BC=C D, &c. and G H=HI=I K, &c. Therefore, whatever be the ratio of A B to G H, the same must be the ratio of B C to H I, of C D to I K, &c., that is, A B:GH::BC:HI::CD:1 K, and so on round the figure. Hence the two polygons are similar. The same reasoning would apply to any other number of sides. 90. —Every regular polygon may be inscribed in a circle and circumscribed about a circle, A polygon is said to be inscribed, when all its vertices are in the circumference, and to be circumscribed, when all its sides are tangents. Let there be a regular polygon A B C D E F (fig. 59). Find F 59 the centre I of a circle (29) to pass through the three points B, C, D. We say the same will pass through all the other vertices. First it will pass through E. Why?-Draw the chords B D and C E. Then, by the definition, the triangles B C D and C D E are equal (53), and if C D were placed upon B C D E would fall upon C D. Accordingly the same circle which passes through B, C, D, will also pass through C, D, E. The same reasoning will apply to F and A, and to any number of vertices. Secondly we say that this polygon may be circumscribed. Draw i H perpendicular to the middle of A B (28) and I G perpendicular to the middle of A F. Describe a circle with the radius I H, and A B will be a tangent (40). Now we say that A F will be a tangent to the same circle. Why? --Because the two right triangles AI H and AIG are equal (59) since they have the hypothenuse A I common, and A H, half of A B=A G, half of A F. Therefore Í H=I G, and A F is a tangent. The same might a cir 92. be proved in like manner of all the other sides. Thus whenever a regular polygon is given, there may be cle circumscribed about it, and a circle inscribed in it, or, in the words of the enunciation, the polygon may be inscribed and circumscribed. 91. We cannot solve the general problem, having a given circle, to inscribe in it a regular polygon of any number of sides, since we have no means of dividing the circumference of a circle into any given number of equal parts. But there are certain particular cases in which the solution is possible. We begin with the square. -Hav ing a given circle to inscribe a squarem. Let the given cirF60 cle be A B C D (fig. 60). Draw two diameters perpen dicular to each other, and join their extremities by chords. A B C D is a square (83), because its sides are equal (53) and its angles are right angles (42). - To inscribe in a given circle a regular hexagon and F61 an equilateral triangle Take the radius AO (fig. 61) in the compasses, and apply it round the circumference. We e say that it will be contained exactly six times, or, in other words, that-the side of an inscribed hexagon is equal to radius Why ?-Because, since A (=B 0, the angle 0 A B=O B A. Then, supposing A B to be the side of a regular hexagon, the angle A O B must be equal to 60°, since the arc A B is a sixth part of the whole circumference. Then the angles 0 A B +0 B A must be equal to 1200 (46), and since they are equal, each must be 60°. Therefore the triangle A O B is equilateral (51), and A B, the side of a hexagon, is equal to the radius A 0. If now we would inscribe an equilateral triangle, it is only necessary to join the alternate vertices A, C, E. Indeed, we may remark generally that when any polygon of an even number of sides is already inscribed, we may always inscribe one of half the number of sides, by joining the alternate vertices. Also, by bisecting the arcs, whether an even number or not, and drawing chords to the half arcs, we may always inscribe one of double the number of sides. 93. - To inscribe in a given circle a regular polygon of ten and one of fift- en sides. First, to inscribe one of ten F62 sides. Divide the radius 0 A (fig. 62) in extreme and mean ratio (82). Let O M be the greater part. Take the chord A B=0 M, and apply it round the circle. We say it will be contained exactly ten times, or, in other |