PROP. D. PROB. Two angles of a triangle being given, to find the third. Draw any straight line CD; at a point therein, as B, make the angle CBA equal to one of the given angles, and the angle ABE equal to the other the remaining angle EBD will be the third angle required; because those three angles (Cor. 13. 1.) are together equal to two right angles. Two angles of a triangle and a side being given, to construct the triangle. The two angles will either be both adjacent to the given side, or the one adjacent and the other opposite: in the latter case, find the third angle (Prop. D.); and the two adjacent angles will thus be known. A Draw the straight line BC equal to the given side; at the point B, make an angle CBA equal to one of the adjacent angles, and at C, an angle BCA equal to the other; the two lines BA, CA, will intersect each other, and form with BC the triangle required; for if they were parallel, the angles B, C, would be together equal to two right angles, and therefore could not belong to a triangle: hence, BAC will be the triangle required. B PROP. F. PROB. Two sides and an angle opposite to one of them being given, to construct the : triangle. This Problem admits of two cases. First. When the given angle is obtuse, make the angle BC'A equal to the given angle; and take C'A equal to that side which is adjacent to the given angle, the arc described from A as a centre, with a radius equal to AB, the other given side, would cut BC on opposite sides of C'; so that only one obtuse angled triangle could be B A formed; that is, the triangle BC'A will be the triangle required. And, if the given angle were right, although two triangles would be formed, yet, as the hypotenuse would meet BC at equal distances from the common perpendicular, these triangles would be equal. Secondly. If the given angle be acute, and the side opposite to it greater than the adjacent side, the same mode of construction will apply: for, making BCA equal to the given angle, and AC equal to the adjacent side; then, from A as centre, with a radius equal to the other given side, describe an arc, cutting CB in B; draw AB, and CAB will be the triangle required. But if the given angle is acute, and the side opposite to it less than the other given side; make the angle CBA equal to the given angle, and take BA equal to the adjacent side; then, the arc described from the centre A, with the radius AC equal to the opposite side, will cut the straight line BC in two points C' and C, lying on the same side of B: hence, there will be two triangles BAC', BAC, either of which will satisfy the conditions of the problem. SCHOLIUM. In the last case, if the opposite side was equal to the perpendicular from the point A on the line BC, a right angled triangle would be formed. And the problem would be impossible in all cases, if the opposite side was less than the perpendicular let fall from the point A on the straight line BC. PROP. G. PROB. To find a triangle that shall be equivalent to any given rectilineal figure. Let ABCDE be the given rectilineal figure. Draw the diagonal CE, cutting off the triangle CDE; draw DF parallel to CE, meeting AE produced, and join CF: the polygon ABCDE will be equivalent to the polygon C D+ The same process may be applied to every other polygon; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at length be found. COR. Since a triangle may be converted into an equivalent rectangle, it follows that any polygon may be reduced to an equivalent rectangle. PROP. H. PROB. A To find the side of a square that shall be equivalent to the sum of two squares. Draw the two indefinite lines AB, AC, perpendicular to each other. Take AB equal to the side of one of the given squares, and AC equal to the other; join BC: this will be the side of the square required. For the triangle BAC being right angled, the square constructed upon BC (47. 1.) is equal to the sum of the squares described upon AB and AC. C D A B SCHOLIUM. A square may be thus formed that shall be equivalent to the sum of any number of squares; for a similar construction which reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. PROP. I. PROB. To find the side of a square equivalent to the difference of two given squares. Draw, as in the last problem, (see the fig.) the lines AC, AD, at right angles to each other, making AC equal to the side of the less square; then, from C as centre, with a radius equal to the side of the other square, describe an arc cutting AD in D: the square described upon AD will be equivalent to the difference of the squares constructed upon AC and CD. For the triangle DAC is right angled; therefore, the square described upon DC is equivalent to the squares constructed upon AD and AC: hence (Cor. 1. 47. 1.), AD2-CD2-AC2. PROP. K. PROB. A rectangle being given, to construct an equivalent one, having a side of a given length. Let AEFH be the given rectangle, and produce one of its sides, as AH, till C B G COR. A polygon may be converted into an equivalent rectangle, having one of its sides of a given length. ELEMENTS OF GEOMETRY. BOOK II. DEFINITIONS. 1. EVERY right angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which are about one of the right angles. "Thus the right angled parallelogram AC is called the rectangle contain"ed by AD and DC, or by AD and AB, &c. For the sake of brevity, "instead of the rectangle contained by AD and DC, we shall simply say "the rectangle AD. DC, placing a point between the two sides of the "rectangle." A. In Geometry, the product of two lines means the same thing as their rectangle, and this expression has passed into Arithmetic and Algebra, where it serves to designate the product of two unequal numbers or quantities, the expression square being employed to designate the product of a quantity multiplied by itself. The arithmetical squares of 1, 2, 3, &c. are 1, 4, 9, &c. 2. In every parallelogram, any of the 66 complements AF, FC, is the gno"mon of the parallelogram AC. This "6 gnomon may also, for the sake of "brevity, be called the gnomon AGK "or EHC." |