Proposition 4. Problem.- To describe a circle about a given regular polygon.

Let ABC be a regular polygon; it is required to describe a circle about it.

Bisect two adjacent angles, A and B, by the lines EA, BE meeting in E; join ED; because the equal angles at A and B are bisected, therefore EAB is equal to EBA, and (I. 23) EA to EB. Also, because in the two triangles EBA, EBD the two sides EB, BA and in

D cluded angle EBA are equal to EB, BD and included angle EBD, the side EA is equal (I. 4) to the side ED. In the same manner all lines drawn from E to the angles of the polygon are equal, and a circle from the centre E, at the distance of any one of them, will pass through the extremities of the others, and will be described about the polygon.

Corollary 1.From this and the last proposition it is evident that, The centres of the circles inscribed in a regular polygon and described about it, are in the same point.

Corollary 2.-Also, that if the radii EA, EB, etc. be produced, they will also bisect the angles of a circumscribed polygon of the same number of sides, which will touch the circle at F, G, etc., the extremities of radii bisecting the angles at the centre. The sides of the two polygons being parallel, the angles will be equal and the polygons similar. Hence, To draw a circumscribed polygon similar to a given inscribed polygon, bisect the angles at the centre, and from the extremities of the radii draw lines touching the circle.


Proposition 5. Problem.-To describe an isosceles triangle having each of the angles at the base double the third angle.

Draw any circle BDE, and in it any radius AB; divide (V. 23) AB in C in extreme and mean ratio; lay down BD equal to AC, and join AD and CD; ABD will be the triangle required.

Because AB is divided in C in extreme and mean ratio, and BD is equal to AC, therefore,

AB : BD :: BD : BC; hence the two triangles ABD, DBC, having a common angle B, have the sides about that angle proportional; they are therefore similar (V. 6); hence the angle DCB is equal to the angle ADB, or ABD, and (I. 23) CD is equal to BD or AC. Because CD is equal to CA, the angle CDA is equal to the angle CAD; but BCD is equal (I. 30) to CAD and CDA; therefore BCD is double CAD; but BCD is equal to ABD or ADB; hence each of the angles ABD, ADB is double BAD.

Proposition 6. Problem.-To inscribe a regular decagon in a given circle.

Let ABC be a given circle ; it is required to describe a regular decagon in it.

Construct (VI. 5) the isosceles triangle DAB on the radius DA, having each of the angles at the base double of the third angle. Because each of the angles DAB, DBA is double ADB, the two



together are four times the angle ADB, and the three angles DAB, DBA, ADB are five times the angle ADB. Therefore the angle ADB is the fifth part of two right angles; that is, the tenth part of four right angles. Therefore, ten such angles as ADB can be exactly placed about the point D (I. 17, Cor. 2). But equal angles stand on equal arcs, and equal arcs are subtended by equal straight lines ; therefore AB is one side of an equilateral decagon inscribed in the circle. And because all triangles ADB, BDE are isosceles and have equal angles at D, their other angles are equal. But these angles make up the angles of the decagon; therefore the decagon is also equiangular; therefore it is regular.

Corollary.- To inscribe a regular pentagon in a given circle.

Join the alternate angles of the decagon, and we have a regular pentagon. For all such triangles as ABE, EFC have their two sides and included angle at B and F equal in each ; hence their third sides are equal, and their angles are equal. But the two angles AEB, CEF, taken from the angle of the decagon, will leave the angles of the pentagon. Therefore the pentagon is equilateral and equiangular.

Proposition 7.

Problem.To inscribe a regular hexagon in a given circle.


Let ABCD be a circle; it is required to describe a regular hexagon in it. Draw

any radius ED. From the centre D and radius DE describe a circle cutting the given circle in A and C.

Join AE, CE, and produce them, and also DE, to meet the circumference in G, F and B; join the




extremities of these lines; the figure formed is a regular hexagon.

Because DA DE = AE, the triangle A ED is equiangular (I. 23, Cor.). Therefore (I. 30, the angle AED is one third of two right angles. For the same reason DEC is one third of two right angles. Therefore the whole angle AEC is two thirds of two right angles, and the angle CEG is (I.15) one third of two right angles. Hence the vertical angles AEF, FEB, BEG are also each equal (I. 17) to one third of two right angles. But equal angles stand on equal arcs, and equal arcs are subtended by equal straight lines ; hence, DA, AF, FB, BG, GC, CD are all equal, and the hexagon is equilateral. It is also equiangular, for the arc DA is equal to the arc FB; add to both the semicircle BCD, and the whole arc BCA is equal to the whole arc FGD. But equal angles stand on equal arcs; hence the angle AFB is equal to the angle DAF. In the same manner all the other angles may be proved equal, and the hexagon is equiangular. It is therefore regular.

Corollary.-A side of a regular hexagon is equal to the radius of the circumscribed circle.

Hence, to construct a hexagon, describe a circle, and lay down the radius six times around it.

Proposition 8. Problem.-To inscribe a regular pentedecagon in a given circle.

Let AB be one side of an equilateral triangle inscribed in a circle (III. 26), and AC one side of a regular penta


gon (VI. 6, Cor.). If the whole circumference were to be divided into 15 equal parts, AB would subtend 5, and AC, 3. Therefore BC must contain 2; bisect (III. 23) BC in D, and layoff BD around the circle; we would have a regular pentedecagon.

Scholium.--If we bisect (III. 23) the arcs of a circle which are subtended by the sides of a square, we may form a regular polygon of 8 sides ; from this, one of 16, and so on. Similarly from a decagon, we can construct polygons of 20, 40, etc., sides; from a hexagon, polygons of 12, 24, etc. ;. from a pentedecagon, figures of 30, 60, etc.



Proposition 9. Theorem.The area of a regular polygon is equal to onehalf the product of its perimeter and apothem.

Let ABC be a polygon and DE its apothem; then the area of ABC is equal to {(AB+BF+FC+CG+GH

+HAED. Join DA, DB, etc.; the area of the triangle DAB is equal to (V. 16, Cor. 2) LAB~ ED. But the apothem is the same for all sides ; hence the area of the whole polygon

= {ABED+1BFx ED+etc.






Proposition 10. Theorem.About a given circle a polygon may be described, which shall differ from a similar inscribed polygon, by a space less than any given space.

Let ABC be a circle; similar polygons may be described

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