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PROP. XVI.

27. THEOREM. The angle of a regular pentagon exceeds a right angle by one-fifth part of a right angle; and is three times as great as the angle contained by any two sides of the figure, which are not adjacent to each other, produced so as to meet.

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angular pentagon, and let any two of its sides, as EA, DB, be produced, so as to meet in H: Any of its exceeds a right-angle by one-fifth part of a right, and is three times as great as the AHB.

About the pentagon ABDCE describe (E. 14. 4.) the circle AECDB; bisect (E. 30.3.) AB, in G, and join C, A and C, B, and C, G and E, G and E, B: And since (hyp. and E. 28. 3.) the circumferences

CE, EA, CD, DB, are equal, CEG=CDG; .. CEG is the semi-circumference of the circle, and (E. 31. 3.) the 4 CEG is a right; also (E. 21. 3.) the AEG=/ ACG; and it is manifest from the demonstration of E. 11. 4. E. 32. 1., that the ACB is the fifth part of two right, and .. that its half, namely the AEG, is the fifth part AEG, which is the excess CEA above a right, is the fifth part of

of a right ; .. the of the

a right 4.

Again, the two opposite AEC, CBA, of the trapezium AECB are (E. 22, 3.) together equal to two right; and (E. 27. 3.) the CBA = 4 BCE;.. the AEC+2 ECB= two right, and .'. (E. 28. 1.) CB is parallel to EA or EH ; .. (E. 29. 1.) the EHD=4CBD, which, since (E, 27. 3.) the three CBD, CBE, and EBA, are equal to one another, is a third part of the

the pentagon ABDCE.

ABD of

28. COR. It is manifest from the demonstration, that the straight line joining the extremities of the first and second side of an equilateral and equiangular pentagon is parallel to the fourth ' side of the figure; the sides being taken in order from any one of them assumed as the first.

PROP. XVII.

29. THEOREM. The square of the side of a regular pentagon, inscribed in a given circle, is equal to

the square of the side of a regular decagon, together with the square of the side of the regular hexagon, both inscribed in that given circle.

Let AECB be the given circle, of which K is

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the centre, and AB the side of a regular decagon inscribed (S. 14. 4. cor. 1.) in, it: Place, in the circle, BC=AB, and join A, C; .'. (S. 14. 4. cor. 3.) AC is the side of a regular pentagon inscribed in the circle, and if KA, KB, and KC be drawn, any one of these lines, as KA, is (E. 15. 4.) the side of a regular hexagon inscribed in the circle: Then AC=KA'+ AB'.

For, in KA take KD = AB, draw BD and produce it to meet the circumference in E;, also, draw KE, KC, CE and CD: Then, it is manifest from S. 14. 4. cor. 1. and E. 10. 4., that the 2 KBD = 2 AKB; but (constr. and E. 8. 1.) the AKB BKC; .. the DBK or EBK = 2 BKC; ... (E. 27. 1.) ED is parallel to KC:

Again, (E. 20. 3.) the 22 EBA; but (E. EDK is equal to the (constr.) to twice the KBE, or to twice the ZEBA; .. the EKDEDK, .. ED=EK; and (E. 15. def. 1.) EKKC; ... ED = KC, and it has been shewn that ED is parallel to KC; .. (E. 33. 1.) EK is equal and parallel to DC, and the figure EKCD is a rhombus ; .. (S. 45. 1.) KD is bisected at right in H, by CE: And since KD is bisected in H and produced to A,

EKA or EKD =

32. 1.) the exterior DKB + KBD, that is,

.. (E. 6. 2.)

KAXAD+DH'=AH2

.. KAXAD+DH2+HCˆ=AHˆ+HC ; but (constr. and S. 14. 4. cor. 1.) KAXAD=AB; and (E. 47. 1.) DH'+HC-DC2, or KC, or KA'; and AH+C=AC;

.. AC-KA+AB.

30. Cor. Hence, if ABC be a given circle,

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and AC, BK two diameters drawn (E. 11. 1.) at right to one another, if KC be (E. 10. 1.) bisected in D, and if from DA there be cut off DE=DB, EK is the side of a regular decagon inscribed in the circle ABC, KB is the side of a regular hexagon inscribed in it, and EB is the side of a regular pentagon inscribed in it.

For (E. 11.2.) EK is equal to that part of KB, the square of which equals the rectangle contained by KB and the remaining part of KB; .. (S. 14. 4. cor. 1.) EK is the side of a regular decagon; and KB (E. 15. 4.) is the side of the regular hexagon, inscribed in the circle ABC; since, .., (constr. and E. 47. 1.) EB’=EK*+KD ̊, EB is (S. 17.4.) the side of a regular pentagon inscribed in the circle ABC.*

PROP. XVIII.

31. PROBLEM. Upon a given finite straight line, to describe an equilateral and equiangular hexagon.

Upon the given straight line, as a base, describe

*This corollary furnishes the best practical method of determining the sides of a regular pentagon, and of a regular de

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