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 pa rallel to AD. For, if A, B, and D, C be joined, it is manifest from the construction, that AD = BC, and AB= DC: therefore (Supp. xvi. 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 angle, as A, of the ABCD be a right angle: The 4s B, C, and D are also right angles. A For, since AD is parallel to BC, and AB meets them, the two interior s A, B are, (E. xxix. 1.) together, equal to two right angles; but (hyp.) the LA is a right angle; therefore the B is also a right angle: And, in the same manner, may the remaining angles, C and D, be shewn to be right angles. PROP. XX. 30. PROBLEM. To trisect a right angle; i. e. to divide it into three equal parts. Let the XAY be a right angle: 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 ▲ ACB; and from A draw (E. xii. 1.) AD perpendicular to BC: The XAY is trisected by the two straight lines AC and AD. For, from C draw (E. xII. 1.) CE perpendicular to AY; then, since the s BAE, AEC, are right angles; therefore (E. xxviii. 1.) AB is parallel to EC; therefore (E. xxix. 1.) ECA CAB = L ACB; because (constr.) the ▲ ACB is equilateral, and (E. v. 1. Cor.) equiangular: Since, therefore, the ACE ACD, and that the 8 D and E are right angles, and AC is common to the two As ADC, AEC; therefore (E. xxvI. 1.) the EAC DAC: Again, since (constr. and E. v. 1. Cor.) the ACB = = ABC, and (constr.) the angles at D are right angles, and that AC=AB; therefore (E. xxvI. 1.) the DACL DAB: But it was shewn that the L EAC-L DAC; .. L EAC= L DAC= L DAB; i. e. the right XAY is trisected by AC and AD. PROP. XXI. 31. PROBLEM. Hence, to trisect a given rectilineal angle, which is the half, or the quarter, or the eighth part, and so on, of a right angle. First, let the given YAZ, be the half of a right angle, and let it be required to trisect it. Draw (E. XI. 1.) from A, AX perpendicular to AY; trisect (Supp. XVIII. 1.) then (Supp. 1. 1.) trisect the half of the YAX. the right XAY; YAZ, which is the But, if the given angle be the quarter of a right angle, its double may be trisected by the former case; and therefore the given angle itself may be trisected by (Supp. 1. 1.). And, by following the same method, it is evident that an angle may be trisected, which is the eighth part, or the sixteenth part, and so on, of a right angle. PROP. XXII. 32. PROBLEM. In the hypotenuse of a rightangled triangle, to find a point, the perpendicular distance of which from one of the sides, shall be equal to the segment of the hypotenuse between the point and the other side. Let ABC be a right-angled triangle, right-angled at C: It is required to find a point in the hypotenuse AB, the perpendicular distance of which from one of the sides, as AC, shall be equal to the segment of the hypotenuse between that point, and BC. Bisect (E. 1x. 1.) the ABC, by BD, and let BD meet AC in D; through D, draw DE (E. xxxI. 1.) parallel to CB: E is the point which was to be found. For, since DE is parallel to CB, the ▲ BDE (E. XXIX. 1.); but (constr.) the CBD=" 4 DBE; . 4 DBE= ‹ BDE; .. (E. vi. 1.) ED = EB; and since (hyp.) the C is a right angle, and that DE is parallel to CB, the ▲ CDE (E. XXIX. 1.) is a right angle; i. e. ED is perpendicular to AC. PROP. XXIII. 33. PROBLEM. In the base of a given acuteangled triangle, to find a point, through which if a straight line be drawn perpendicular to one of the sides, the segment of the base, between that side and the point, shall be equal to the segment of the perpendicular, between the point and the other side produced. Let ABC be the given acute-angled triangle: It is required, to find, in the base BC, a point through which if a perpendicular be drawn to AB, the segment of the base, between that point and the point B, shall be equal to the segment of the perpendicular between that same point and AC produced. Draw (E. XI. 1.) from B, BY perpendicular to AB; bisect (E. Ix. 1.) the CBY by BD, meeting AC produced in D; through D, draw (E. xxxi. 1.) |