2. Prove that BD is the side of a regular decagon inscribed in the circle BDE.
3. If DB, DE, EF be consecutive sides of a regular decagon inscribed in a circle, prove BF − BD = radius of circle.
4. If E be the second point of intersection of the circle ACD with BDE, DE is equal to DB; and if AE, BE, CE, DE be joined, each of the triangles ACE, ADE is congruent with ABD.
5. AC is the side of a regular pentagon inscribed in the circle ACD, and EB the side of a regular pentagon inscribed in the circle BDE.
6. Since ACE is an isosceles triangle, EB2 − EA2 = AB.BC—that is = BD2; therefore EB2 − BD2 = EA2—that is, the square of the side of a pentagon inscribed in a circle exceeds the square of the side of the decagon inscribed in the same circle by the square of the radius.
PROP. XI.—Problem.
To inscribe a regular pentagon in a given circle (ABCDE).
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.