Sol.—Construct an isosceles triangle [x.], having each base angle double the vertical angle, and inscribe in the given circle a triangle ABD equiangular to it. Bisect the angles DAB, ABD by the lines AC, BE. Join EA, ED, DC, CB; then the figure ABCDE is a regular pentagon. Dem.—Because each of the base angles BAD, ABD is double of the angle ADB, and the lines AC, BE bisect them, the five angles BAC, CAD, ADB, DBE, EBA are all equal; therefore the arcs on which they stand are equal; and therefore the five chords, AB, BC, CD, DE, EA are equal. Hence the figure ABCDE is equilateral. Again, because the arcs AB, DE are equal, adding the arc BCD to both, the arc ABCD is equal to the arc BCDE, and therefore [III. xxvii.] the angles AED, BAE, which stand on them, are equal. In the same manner it can be proved that all the angles are equal; therefore the figure ABCDE is equiangular. Hence it is a regular pentagon. Exercises. 1. The figure formed by the five diagonals of a regular pentagon is another regular pentagon. 2. If the alternate sides of a regular pentagon be produced to meet, the five points of meeting form another regular pentagon. 3. Every two consecutive diagonals of a regular pentagon divide each other in extreme and mean ratio. 4. Being given a side of a regular pentagon, construct it. 5. Divide a right angle into five equal parts. PROP. XII.—Problem. To describe a regular pentagon about a given circle (ABCDE). Sol.—Let the five points A, B, C, D, E on the circle be the vertices of any inscribed regular pentagon: at these points draw tangents FG, GH, HI, IJ, JF: the figure FGHIJ is a circumscribed regular pentagon. Dem.—Let O be the centre of the circle. Join OE, OA, OB. Now, because the angles A, E of the quadrilateral AOEF are right angles [III. xviii.], the sum of the two remaining angles AOE, AFE is two right angles. In like manner the sum of the angles AOB, AGB is two right angles; therefore the sum of AOE, AFE is equal to the sum of AOB, AGB; but the angles AOE, AOB are equal, because they stand on equal arcs AE, AB [III. xxvii.]. Hence the angle AFE is equal to AGB. In like manner the remaining angles of the figure FGHIJ are equal. Therefore it is equiangular.