45. To bisect a parallelogram by a line drawn from a point in one of its sides. 46. To bisect a given triangle by a line drawn from a point in one of its sides. 47. If three straight lines intersecting in the same point make the six angles thus formed equal, and a point be taken in one of the angles, the sum of the perpendiculars from this point upon the sides that contain the angle, will be equal to a perpendicular from the point on the third line. 48. Given a point and three lines, two of which are parallel, to find a point in each of the parallels that shall be equidistant from the given point, and such, that the line joining them shall be parallel to the other given line. 49. A line joining the middle points of two sides of a triangle, is parallel to the base, and equal to the half of it. 50. The quadrilateral formed by joining the successive middle points of the sides of a given quadrilateral, is a parallelogram. ! 51. If any parallelogram be described on the base of a triangle, and other two parallelograms on its two sides, such that their sides opposite to those of the triangle shall pass through the angular points of the former, the first parallelogram shall be equal to the sum or difference of the other two, according as they both lie without the triangle, or one of them upon it. 52. If one of the oblique angles of a triangle be divided into a number of equal parts, the dividing lines will divide the opposite side unequally ; the segments nearer to the perpendicular being less than the more remote. 53. Any three lines being drawn making equal angles with the three sides of any triangle towards the same parts, and meet ing one another, will form a triangle equiangular to the original triangle. SECOND BOO K. DEFINITIONS. * 1. If a straight line be divided or cut at any point, the point is called a point of section. · 2. The distance of the point of section from the middle of the line, is called the mean distance. 3. If the point of section lie between the extremities of the line, it is said to be cut internally; and if beyond one of the extremities, it is said to be cut externally. 44. The distances from a point of section of a line to its extremities, are called segments; which are said to be internal or external, according as the line is cut internally or externally. 5. The rectangle under, or contained by, two lines, is a rectangle, of which these lines, or lines equal to them, are two adjacent sides. The rectangle under, or contained by, two lines, as AB and BC, is sometimes concisely expressed thus, AB · BC; or if A and B denote the lines, by A.B. So the rectangle under, or contained by, a line, which is the sum of two lines A and B, and a third line C, is expressed thus (A + B) C; that under, or contained by, the excess of A above B, and another line C, by (A - BC; and that under a line equal to A + B, and another equal to A – B, by (A + B) (A – B). 6. When a line is cut into two segments, so that the rectangle under the whole line and one of the segments is equal to the square of the other segment, it is said to be cut medially, or in inedial section. 7. Any of the parallelograms about a diagonal of any parallelogram, together with the two complements, is called a gnomon. Thus the parallelogram HG, together with A I the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon. PROPOSITION I. THEOREM. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Given the two straight lines A and BC, and let BC be divided into any parts in the points D, E; to prove that the rectangle contained by the straight lines A and BC, is equal to the rectangle B D E C contained by A and BD, together with that contained by A and DE, and that contained by A and EC. (Const.) From the point B draw (I. 11) GL BF at right angles to BC, and make K L H BG equal to A (I. 3); and through G Fl and draw GH parallel to BC (I. 31); A through D, E, and C draw (I. 31) DK, EL, and CH parallel to BG. (Dem.) Then the rectangle BH is equal to the rectangles BK, DL, and EH; and BH is contained by A and BC, for it is contained by GB and BC, and GB is equal to A; and BK is contained by A and BD, for it is contained by GB and BD, of which GB is equal to A; and DL is contained by A and DE, because DK, that is, BG (I. 34), is equal to A; and in like manner the rectangle EH is contained by A and EC; therefore the rectangle contained by A and BC, is equal to the several rectangles contained by A and BD, by A and DE, and by A and EC. PROPOSITION II. THEOREM. If a straight line be divided into any two parts, the rectangles contained by the whole line and each of its parts are together equal to the square on the whole line. Given the straight line AB divided into any two parts in the point C; to prove that the rectangle contained by AB and BC, together with the rectangle contained by AB and AC, shall be equal to the square on AB. (Const.) Upon AB describe (I. 46) the square ACE ADEB; and through C draw CÉ parallel to AD or BE (I. 31); (Dem.) then AE is equal to the rectangles AF and CE and AE is the square of AB; and AF is the rectangle contained by BA and AC; for it is contained by DE E DA and AC, of which' AD is equal to AB; and CE is contained by AB and BC, for BE is equal to AB; therefore the rectangle contained by AB and AC, together with the rectangle AB · BC, is equal to the square on AB. PROPOSITION III. THEOREM. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the square on that part, together with the rectangle contained by the two parts. Given the straight line AB divided into two parts in the point C; to prove that the rectangle AB · BC is equal to the square on BC, together with the rectangle AC · CB. (Const.) Upon BC describe (I. 46) the square CDEB, and produce ÉD to F, and through A draw AF parallel to CD or BE (I. 31). (Dem.) Then the rectangle AE is equal to the rectangles CE and AD. Now, AE is the rectangle contained by AB and BC, for it is contained by AB and, BE, of which BE is equal to BC; and CE is the square on BC; and AD is the rectangle contained by AC and CB, for CD is equal to CB; therefore the rectangle AB · BC is equal to the square on BC, together with the rectangle AC.CB. PROPOSITION IV. THEOREM. If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the squares on the two parts, together with twice the rectangle contained by the parts. Given the straight line AB divided into any two parts in C; to prove that the square on AB is equal to the squares on AC and CB, together with twice the rectangle contained by AC and CB. (Const.) Upon AB describe the square ADEB (I. 46), and join BD, and through C draw CGF parallel to AD or BE (I. 31), and through G draw A_ C. HK parallel to AB or DE. (Dem.) And because CF is parallel to AD, and BD falls upon them, the exterior angle CGB is equal to the interior and opposite angle ADB (I. 29); but ADB is equal to the angle ABD (. 5), because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle CBG; and therefore the side BC is equal to the side CG (I. 6); but CB is equal also to GK (I. 34), and CG to BK; wherefore the figure CGKB is equilateral. It is likewise rectangular; for the angle CBK being a right angle, the other angles of the parallelogram CGKB are also right angles (I. 46, Cor.); wherefore CGKB is a square, and it is upon the side CB. For a like reason, HF is also a square, and it is upon the side HG, which is equal to AC; therefore HF and CK are the squares on AC and CB. And because the complement AG is equal (I. 43) to the complement GE, and that AG is the rectangle contained by AC and CG, that is, by AC and CB; GE is also equal to the rectangle AC · CB; wherefore AG, GE, are together equal to twice the rectangle AC · CB.. And HF, CK are the squares on AC and CB; wherefore the four figures HF, CK, AG, and GE are equal to the squares on AC and CB, and to twice the rectangle AC · CB. But HF, CK, AG, and GE make up the whole figure ADEB, which is the square on AB; therefore the square on AB is equal to the squares on AC and CB, and twice the rectangle AC • CB. Cor.-From the demonstration, it is manifest that the parallelograms about the diagonal of a square are likewise squares. PROPOSITION V. THEOREM. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Given the straight line AB, divided into two equal parts in the point C, and into two unequal parts at the point D; to prove that the rectangle AD · DB, together with the square on CD, is equal to the square on CB. (Const.) Upon CB describe the square CEFB (I. 46); join BE, and through D draw DHG parallel to CE or BF (I. 31); and through H draw KLM parallel to CB or EF; and also through A A draw AK parallel to CL or BM. (Dem.) And because the complement CH is equal to the complement HF (I. 43), to each of these add DM; therefore the whole CM is equal to the whole E GB DF; but CM is equal to AL (I. 36), because AC is equal to CB; therefore also AL is equal to DF. To each of these add CH, and the whole AH is equal to DF and CH; but AH is the rectangle contained by AD and DB, for DH is equal to DB (II. 4, Cor.); and DF, together with CH, is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD • DB. To each of these add LG, which is equal to the square on CD; therefore the gnomon CMG, together with LG, is equal to the rectangle |