PROP. XLV. PROBLEM. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. the figure It is required to describe a parallelogram CONSTRUCTION.-1. Join BD and describe the parallelogram FH the triangle ABD, and having the angle FKH the angle E (I. 42). 2. Apply to the straight line GH the parallelogram GM equal to the triangle DBC, and having the angle GHM the angle E (I. 44). Then it is to be proved that FM is a parallelogram the figure ABCD, and having the angle FKM = the angle E. PROOF.-Because each of the angles FKH and GHM the angle E (cons.), therefore the angle FKH = the angle GHM (ax. 1); add to each the angle GHK, then the angles FKH and GHK = the angles GHM and GHK (ax. 2). But the angles FKH and GHK = two right angles (I. 29); therefore the angles GHM and GHK = two right angles, and therefore KH is in the same straight line with HM (I. 14). 2. Next, because the lines FG and KM are parallel (cons.), and GH meets them, therefore the alternate angle FGH the alternate angle GHM (I. 29); add to each the angle LGH, then the angles FGH and LGH = the angles GHM and LGH (ax. 2). But the angles MHG and LGH = two right angles (I. 29); therefore the angles HGF and LGH = two right angles, and therefore FG is in the same right line with GL (I. 14). 3. Next, because KF is parallel to HG, and HG to ML (cons.), therefore KF is parallel to ML (I. 30), and KM and FL are also parallel (cons.); therefore the figure FKML is a parallelogram (def. 34). Again, because the parallelogram FH the triangle the triangle DBC (cons.), therefore the parallelogram FM = the figure ABCD, and it has the angle FKM = the angle E (cons.) ABD (cons.), and the parallelogram GM Therefore, it is proved, as required, that FM is a parallelogram = the figure ABCD, and having an angle FKM = the angle E. Q. E. F. Corollary.-A parallelogram may be described equal to a given rectilineal figure, of any number of sides, and having one of its angles equal to a given angle. N.B.-The learning of this Proposition will be simplified if it be observed— 1. That the given figure is divided into triangles. 2. That a parallelogram is made equal to each triangle, the first according to Euclid I. 42, the second, and more if there be any, according to I. 44. 3. That the first part of the proof is to shew that KH and HM are in one and the same straight line. 4. That the second part is to shew that FG and GL are in one and the same straight line. PROP. XLVI. PROBLEM. To describe a square upon a given straight line. Let AB be the given straight line. It is required to describe a square upon AB. CONSTRUCTION.-1. From A draw AC at right angles to and greater than AB (I. 11), cutting off AD = AB (I. 3). 2. Through D draw DE parallel to AB: and through B draw BE parallel to AD (I. 31). Then it is to be proved that The figure ABED is a square upon AB. PROOF.-1. The figure ABED is equilateral. Because DE is parallel to AB (cons.), and BE is parallel to AD (cons.), therefore the figure ABED is a parallelogram (def. 34), and therefore, also, AB = DE and AD = BE (I. 34). But also AB AD (cons.), therefore the four sides AB, BE, ED, and DA each other, and therefore the figure ABED is equilateral, which was first to be shewn. 2. The figure ABED is rectangular. Next, because the lines AB and DE are parallel (cons.), and the line AD falls upon them, therefore the two interior angles BAD and ADE= two right angles (I. 29). But because BAD is a right angle (cons.), therefore ADE is a right angle (ax. 3); and because the opposite angles of parallelograms are equal to each other (I. 34), therefore the angles BED and ABE are also right angles, and therefore each of the angles of the figure a right angle, and the figure ABED is rectangular. Which was next to be shown. Hence, the figure ABED, being equilateral and rectangular is a square (def. 30), and it is described upon AB (cons.). Therefore, it is proved, as required, that The figure ABED is a square upon AB. Exercises. Q. E. F. 1. On a given straight line MN construct a square, with perp. NP drawn from N. 2. About MO, as a diagonal, construct a square MNOP. E In any right-angled triangle, the square which is described upon the side subtending the right angle equals the sum of the squares described upon the sides containing the right angle. Let ABC be a right-angled triangle having the right angle BAC. Then it is to be proved that The square described upon BC, the side subtending―i.e. opposite to the right angle BAC = the sum of the squares described upon BA and AC, the sides containing that angle. CONSTRUCTION.-1. On BC describe the square BDEC: on BA describe the square BAGF: and on AC describe the square AHKC (I. 46). 2. Through A draw AL parallel to BD or CE (I. 31), and join AD and FC. PROOF.-Because the adjacent angles BAC and BAG = two right angles (hyp. and cons.), therefore CAG is one and the same straight line (I. 14). |