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

<DAH > the

PROP. XXXIII.

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

Let AB be the given straight line: It is re

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 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

L

DEF=

▲ CAB, and ▲ DFE=2 CBA ... (S. 26. 1.) ≤ EDF =ACB; but (E. 5. 1. cor. and constr.) the A CAB is equiangular; .. the ADEF is equiangular; and .. (E. 6. 1. cor.) it is, also, equilate

1

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 = 4 DAE,.. ¿EDA=¿ 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

D two given rectilineal angles: It is required to

describe a triangle, which shall have its perimeter equal to AB, two of its angles equal to C and D, each to each, and its third angle equal to an angle, which, together with C and D, makes up two right angles.

At the point A, in AB, make (E. 23. 1.) the BAE= C; and at the point B make the ABE

2D; .. (S. 26. 1.) the AEB is equal to the third of the A which is to be described: Bisect (E. 9. 1.) the EAB, EBA, by AF and BF, which meet in F; and through F draw (E. 31. 1.) FG parallel to EA, and FH parallel to EB: Then is FGH the A, which was to be described.

For, since (constr.) FG is parallel to EA, and FA meets them, .. (E. 29. 1.) the ▲ EAF=2 AFG; but (constr.) the EAF FAG; .. the FAG= AFG; .. (E. 6. 1.) FGGA; and, in the same manner, it may be shewn that FH=HB; ... FG + GH + HF⇒ AG + GH +HB; i. e. the perimeter of the ▲ FGH is equal to the given straight line AB.

Again, because FG is parallel to EA, and FH is parallel to EB, ... (E. 29. 1.) the ▲ FGH=4 EAB, and FHG EBA; but (constr.) the < EABC, and the EBA= 4D; ... also, the FGH C, and the FHB=D;

(S. 26. 1.) the GFH is equal to the third of the A, which was to be described; .. the A FGH, the perimeter of which has been shewn to be equal to the given straight line AB, is the A which was to be described.

PROP. XXXV.

49. THEOREM. If, in the sides of a given square, at equal distances from the four angular points, four other points be taken, one in each side, the figure contained by the straight lines which join them, shall also be a square.

Let ABCD be a given square; in the sides

AB, BC, CD, DA, let the four points E, H, G, F be taken, so that E is at the same distance from A that H is from B, that G is from C, and F from D; and let E, H, and H, G, and G, F, and F, E, be joined: The figure EFGH is a square.

For, since (E. 30. def. 1.) all the sides of the given square ABCD are equal, and that (hyp.) AE= BHDF, it is manifest that the two FAE, EBH have the two sides FA, AE equal to the two EB, BH, each to each, and (E. 30. def. 1.) the AB; .. (E. 4. 1.) the ▲ AFE BEH; and FEEH: And, in the same manner, it

may

be shewn that EHHG=GF; .. the figure EFGH is equilateral.

Again, since, as hath been proved, the AFE = BEH, .. the AFE+AEF=▲ BEH +AEF; but, since the A is a right 4,.. (E. 32. 1.) AFE+ 2 AEF= a right; .. also, < BEH + 2 AEF=a right ; but (E. 15.1. Cor. 2.) < BEH+AEF + 2 HEF=two right ; ..

the HEF is a right angle; and, in the same manner, may the remaining of the figure EFGH, which has been shewn to be equilateral, be proved to be right; ... (E. 30. def. 1.) EFGH is a

square.

PROP. XXXVI.

50. THEOREM. If the opposite angles, of a quadrilateral figure be equal to each other, the figure shall be a parallelogram.

Let AB be a quadrilateral figure, having the

A

C

D

B

angle A equal to the opposite angle B, and the angle C to the opposite angle D: The figure ADBC is a parallelogram.