DEF, is equal to the hypotenuse AB, of the A ABC. For, since AC=DG, and the two ACB, ABC, of the ▲ ABC, are equal to the two L, DGE, DEG, of the A DEG, each to each, ... (E. 26.1.) DE AB. 25. THEOREM. If the sides of any given equilateral and equiangular figure of more than four sides, be produced so as to meet, the straight lines, joining their several intersections, shall contain an equilateral and equiangular figure, of the same number of sides as the given figure. angular figure, of more than four sides; let the sides, produced, meet in the points G, H, I, K, ` L, M; and let those points of intersection be joined: Then is GHIKLM an equilateral and equiangular figure, of the same number of sides as ABCDEF. For, since (hyp.) the A, B, C, D, E, F are all equal, the MAB, GBC, HCD, IDE, KEF, LFA are all (E. 13. 1. and E. 6. 1.) isosceles, and any two of them have their equal, each to each; ... since (hyp.) BA = AF, and that the MAB, MBA are equal to the LFA, LAF, each to ; each, the side MA of the A MAB the side LA (E. 26. 1.) of the A LAF; and in the same manner it may be shewn that MB=GB, GC=CH, HD=DI, IE=EK, and KF=FL: But, because the of the figure ABCDEF are (hyp.) equal, .. (E. 15. 1.) the LAM, MBG, GCH, HDI, IEK, KFL, are all equal to one another ... (E. 4. 1.) the sides LM, MG, GH, HI, IK and KL are all equal, as are also the of the ALAM, MBG, GCH, HDI, IEK, and KFL, each to each: And the AMB, BGC, CHD, DIE, EKF, and FLA have been shewn to be equal to one another: Wherefore the figure GHIKLM is equilateral and equiangular; and it is manifest that it has the same number of sides as the figure ABCDEF. PROP. XVIII. 26. THEOREM. If two opposite sides of a quadrilateral figure be equal to one another, and the two remaining sides be also equal to one another, the figure is a parallelogram. Let any two opposite sides, as AB, DC, of the quadrilateral figure ABCD, be equal to one another, and let the two remaining sides, AD, BC, be, also, equal to one another: The figure ABCD is a . For, join D, B: Then since the two sides AD, DB, of the A ADB, are equal to the two sides CB, BD, of the ▲ CBD, and that the base AB is equal (hyp.) to the base DC, ... (E. 8. 1.) the ▲ ADB=2 DBC; and (E. 4. 1.) the ▲ ABD= ▲ BDC; ...(E. 27. 1.) AD is parallel to BC, and AB is parallel to DC; i. e. the figure ABCD is a . 27. COR. 1. Hence may be deduced a practical method of drawing a straight line, through a given point, parallel to a given straight line. For, let it be required to draw through the given point B, a straight line parallel to AD: From any point A in AD, as a centre, and at any distance, describe a circle cutting AD in D; and from B as a centre, at the same distance, describe another circle; lastly, from D as a centre, at a distance equal to that of A, B, describe another circle, cutting the circle last described in C; join B, C. BC is parallel to AD. For, if A, B and D, C be joined, it is manifest from the construction, that AD=BC, and AB=DC: ... (S. 16. 1.) BC is parallel to AD. 28. COR. 2. A rhombus is a parallelogram. PROP. XIX. 29. THEOREM. Every parallelogram which has one angle a right angle, has all its angles right angles. Let one, as A, of the ABCD be a right angle: The B, C, and D are also right angles. For, since AD is parallel to BC, and AB meets them, the two interior A, B are, (E. 29. 1.) together, equal to two right; but (hyp.) the A is a right; .. the B is also a right : And, in the same manner, may the remaining 4, C and D, be shewn to be right . PROP. XX. 30. PROBLEM. To trisect a right angle; i. e. to divide it into three equal parts. Let the XAY be a right : It is required to trisect it; i. e. to divide it into three equal parts. In AX take any point B; upon AB describe (E. 1. 1.) the equilateral A ACB; and from A draw (E. 12. 1.) AD 1 to BC: The XAY is trisected by the two straight lines AC and AD. For, from C draw (E. 12. 1.) draw CE to AY; then, since the BAE, AEC, are right .. (E. 28. 1.) AB is parallel to EC; ... (E. 29. 1.) ▲ ECA = 2 CAB=▲ ACB; because (constr.) the AACB is equilateral, and (E. 5. 1. cor.) equiangular: Since, ..., the 2 ACE = 2 ACD, |