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Again, since (constr.) AE= DE, the EAD =EDA (E. 5. 1.); but (constr.) AED is a right; each of the EAD, EDA, is half a right; and, in the same manner, may each of the EDB, DBE, CBE, BCE, ECA, EAC, be shewn to be half a right ; .. all the of the figure ADBC are right; and it has been proved that all its sides are equal; .. (E. 30. def. 1.) ADBC is a square.

PROP. XXXI.

44. THEOREM. If either of the acute angles of a given right-angled triangle be divided into any number of equal angles, then, of the segments of the base, subtending those equal angles, the nearest to the right angle is the least; and, of the rest, that which is nearer to the right angle is less than that which is more remote.

Let ACB be a right-angled A, right-angled at C,

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and let the acute 4 BAC be divided into any num

ber of equal, CAD, DAE, EAB, &c.; then is CD the least of the segments of the base subtending those equal, and of the rest DE< EB; and

so on.

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For, at the point D in AD make (E. 23. 1.) ADF=2 ADC: And since, also, the ▲ CAD DAE (hyp.) and AD common to the ACD, AFD, .. (E. 26. 1.) DFDC : But (E. 19. 1. and E. 32. 1.) DE> DF; '.. DE> DC; i. e. DC <DE.

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Again, at the point E, in AE, make the ▲ AEK = AED; and it may, in like manner, be shewn that EKED: But (E. 16. 1.) 4 BKE> < AEK; .. ¿BKE > < AED; and ▲ AED> ▲ ABE; much more then is BKE > < EBK; .. (E. 19. 1.) BE > EK or ED; i. e. ED < EB.

And in the same manner may EB be shewn to be less than the next segment that is more remote from C; and so on.

45. COR. It is manifest, from the demonstration, that if any three straight lines AB, AE, AD, be drawn to the given straight line XC from a given point A, without it, so that the BAE= <EAD, the segment BE, of XC, which is the further from the perpendicular AC, shall be greater than the segment ED, which is the nearer to AC.

PROP. XXXII.

46. THEOREM. If either angle at the base of a

triangle be a right angle, and if the base be divided into any number of equal parts, that which is adjacent to the right angle shall subtend the greatest angle at the vertex; and, of the rest, that which is nearer to the right angle shall subtend, at the vertex, a greater angle than that which is more remote.

Let ACB be a right-angled A, right-angled at

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C, and let the base BC be divided into any number of equal parts CD, DH, HG, &c.: Of these segments DC shall subtend the greatest at the vertex A; and of the rest DH shall subtend, at A, a greater than HG; and so on.

For, join A, D, and A, H, and A, G, &c.; also, at the point A, in DA, make (E. 23. 1.) the ▲ DAE CAD: Then (S. 31. 1.) ED> DC; but (hyp.) DC=DH; .. ED>HD, and it is manifest that the EAD HAD; but (constr.) LEAD CAD; .. 2 CAD> <DAH:

And, in the same manner, it may be shewn, by

the help of the corollary to S. 30. 1. that the <DAH > the HAG; and so on.

PROP. XXXIII.

47. PROBLEM. To trisect a given finite straight line.

Let AB be the given straight line: It is re

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quired to divide it into three equal parts.

Upon AB describe (E. 1. 1.) the equilateral A CAB; bisect (E. 9. 1.) the two equal Д A and B, by the straight lines AD and BD, which meet in D; and from D draw (E. 31. 1.) DE parallel to CA, and DF parallel to CB: Then are AE, EF and FB equal to one another.

For, since (E. 29. 1. and constr.) the ▲ DEF= ▲ CAB, and DFE CBA .. (S. 26. 1.) ≤ EDF L =ACB; but (E. 5. 1. cor. and constr.) the ▲ CAB is equiangular; .. the ▲DEF is equiangular; and.. (E. 6. 1. cor.) it is, also, equilate

ral; so that DE and DF are, each of them, equal to EF.

Again, since (E. 29. 1. and constr.) the ▲ EDA =<DAC; and that (constr.) the DAC= 2 DAE,.. EDA=2 DAE; =¿ DAE; .. (E. 6. 1.) AE DE; but DE has been proved to be equal to EF; .. AE EF; and in the same manner, EF may be shewn to be equal to FB; ... AB has been divided into the three equal parts AE, EF, and FB.

48. PROBLEM.

PROP. XXXIV.

To describe a triangle which shall have its three sides, taken together, equal to a given finite straight line, and its three angles equal to three given angles, each to each; the three given angles being together equal to two right angles.

Let AB be a given finite straight line, and C and

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D two given rectilineal angles: It is required to

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