PROBLEM V. Through a given point, in a given angle, to draw a straight line so that the segments between the point and the sides of the angle shall be equal. Let BCD be the given angle, and A the given point. C D but, FE is equal to EC; hence, FA is equal to AD. PROBLEM VI. To construct a triangle equal to a given polygon. Let ABCDE be the given polygon. Draw CA; produce EA, and draw BG parallel to CA; draw the line CG. Then the triangles BAC and GAC have the common base AC, and because their vertices B and G lie in the same line BG parallel to the base, their altitudes are equal, and consequently, the triangles are equal: hence, the polygon GCDE is equal to the polygon ABCDE. A E Again, draw CE; produce AE and draw DF parallel to CE; draw also CF; then will the triangles FCE and DCE be equal: hence, the triangle GCF is equal to the polygon GCDE, and consequently, to the given polygon. In like manner, a triangle may be constructed equal to any other given polygon. 78 PROBLEM VII. To construct a square equal to a given triangle. Let ABC be the given triangle, AD its altitude, and BC its base. square then will this be the square required. For, from the construction, Scholium. By means of Problems VI. and VII., a square may be constructed equal to any given polygon. PROBLEM VIII. On a given straight line, to construct a polygon similar to a In like manner, construct the triangle FHI similar to ACD, and FIK similar to ADE; then will the polygon FGHIK be similar to the polygon ABCDE (P. XXVI., C. 3). PROBLEM IX. To construct α square equal to the sum of two given given squares. 1o. Let A and B be the sides of the given squares, and let A be the greater. Construct a right angle CDE; make DE equal to A, and DC equal to B; draw CE, and on it D A B E construct a square: this square will be equal to the sum a square this square will be equal to Scholium. A polygon may be constructed similar to either of two given polygons, and equal to their sum or difference. For, let A and B be homologous sides of the given polygons a side, homologous to A or B, construct a polygon similar 13 BOOK V. REGULAR POLYGONS.-AREA OF THE CIRCLE. DEFINITION. 1. A REGULAR POLYGON is a polygon which is both equilateral and equiangular. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar. Let ABCDEF and abcdef be regular polygons of the same number of sides: then will they be similar. angles, divided by the number of angles (B. I., P. XXVI., C. 4); and further, the corresponding sides are proportional, because all the sides of either polygon are equal (D. 1): hence, the polygons are similar (B. IV., D. 1); which was to be proved. PROPOSITION II. THEOREM. The circumference of a circle may be circumscribed about any regular polygon; a circle may also be inscribed in it. 1o. Let ABCF be a regular polygon: then can the circumference of a circle be circumscribed about it. For, through three consecutive vertices A, B, C, describe the circumference of a circle (B. III., Problem XIII., S.). Its centre 0 will lie on PO, drawn perpendicular to BC, at its middle point P; draw OA and OD. H G F D Let the quadrilateral OPCD be turned about the line OP, until PC falls on PB; then, because the angle C is equal to B, the side CD will take the direction BA; and because CD is equal to BA, the vertex D, will fall upon the vertex 4; and consequently, the line OD will coincide with OA, and is, therefore, equal to it: hence, the circumference which passes through A, B, and C, will pass through D. In like manner, it may be shown that it will pass through all of the other vertices: hence, it is circumscribed about the polygon; which was to be proved. . A circle may be inscribed in the polygon. For, the sides AB, BC, &c., being equal chords 0 the circumscribed circle, are equidistant from the centre O hence, if a circle be described from O as a centre, with OP as a radius, it will be tangent to all of the sides or the polygon, and consequently, will be inscribed in it; which was to be proved. |