PROP. XLV. PROB. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a ginen rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe (42. 1.) the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH (44. 1.) apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. And because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal (29. 1.) to two right angles; therefore also KHG, GHM are equal to two right angles: and because at the point H in the straight lines GH, the two straight lines KH, HM, upon the opposite sides of GH, make the adjacent angles equal to two right angles, KH is in the same straight line (14. 1.) with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (29. 1.); add to each of these the angle HGL: therefore the angles MHG, HGL, are equal to the angles HGF, HGL: But the angles MHG, HGL, are equal (29. 1.) to two right angles; wherefore also the angles HGF, HGL, are equal to two right angles, and FG is therefore in the same straight line with GL. And because KF is parallel to HG, and HG to ML, KF is parallel (30. 1.) to ML; but KM, FL are parallels: wherefore KFLM is a parallelogram. And because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying (44. 1.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. Let AB be the given straight line: it is required to describe a square upon AB. C D E From the point A draw (11. 1.) AC at right angles to AB; and make (3. 1.) AD equal to AB, and through the point D draw DE parallel (31. 1.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equal (34. 1.) to DE, and AD to BE; but BA is equal to AD: therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelograin ADEB is equilateral; it is likewise rectangular; for the straight line AD meeting the parallels, AB, DE, makes the angles BAD, ADE equal (29. 1.) to two right angles; but BAD is a right angle ; therefore also ADE is a right angle now the opposite angles of parallelograms are`equal (34. 1.); therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB. A B COR. Hence every parallelogram that has one right angle has all its angles right angles. PROP. XLVII. THEOR. In any right angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right angled triangle having the right angle BAC; the square described upon the side BC is equal to the squares described upon BA, AC. On BC describe (46. 1.) the square BDEC, and on BA, AC the squares GB, HC; and through A draw (31. 1.) AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right angle (25. def.), the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the same straight line (14. 1.) with AG; for the same reason, AB and AH are in the same straight line. Now because the angle DBC is equal to the angle FBA, each of them being a right angle, adding to each the angle ABC, the whole angle DBA will be equal (2. Ax.) to the whole FBC; and because the two sides AB, BD, are equal to the two FB, BC each to each, and the angle DBA equal to the angle FBC, therefore the base AD is equal (4. 1.) to the base FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double (41. 1.) of the triangle ABD, because they are upon the same base, BD, and between the same parallels, BD, AL; and the square GB is double of the triangle BFC because these also are upon the same base FB, and between the same parallels FB, GC. Now the doubles of equals are equal (6. Ax.) to one another; therefore the parallelogram BL is equal to the square GB: And in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CI, is equal to the square HC. Therefore, the whole square BDEC is equal to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. COR. 1. Hence, the square of one of the sides of a right angled triangle is equivalent to the square of the hypotenuse diminished by the square of the other side; which is thus expressed: AB2=BC2—AC2. COR. 2. If AB=AC; that is, if the triangle ABC be right angled and isosceles; BC2=2AB2=2AC2; therefore, BC=AB√2. COR. 3. Hence, also, if two right angled triangles have two sides of the one, equal to two corresponding sides of the other; their third sides will also be equal, and the triangles will be identical. If the PROP. XLVIII. THEOR. square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle. If the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle. D From the point A draw (11. 1.) AD at right angles to AC, and make AD equal to BA, and join DC. Then because DA is equal to AB, the square of DA is equal to the square of AB; To each of these add the square of AC; therefore the squares of DA, AC are equal to the squares of BA, AC. But the square of DC is equal (47. 1.) to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore, the square of DC is equal to the square of BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC likewise equal to the base BC, the angle DAC is equal (8. 1.) to the angle BAC; But DAC is a right angle; therefore also BAC is a right angle. B ADDITIONAL PROPOSITIONS. PROP. A. THEOR. A perpendicular is the shortest line that can be drawn from a point, situated without a straight line, to that line: any two oblique lines drawn from the same point on different sides of the perpendicular, cutting off equal distances on the other line, will be equal; and any two other oblique lines, cutting off unequal distances, the one which lies farther from the perpendicular will be the longer. If AB, AC, AD, &c. be lines drawn from the given point A, to the indefinite straight line DE, of which AB is perpendicular; then shall the perpendicular AB be less than AC, and AC less than AD, and so on. For, the angle ABC being a right one, the angle ACB is acute, (17. 1.) or less than the angle ABC. But the less angle of a triangle is subtended by the less side (19.1.) therefore, the side AB is less than the side AC. Again, if BC BE; then the two oblique lines AC, AE, are equal. For the side AB is common to the two triangles ABC, ABE, and the contained angles ABC and ABE equal; the two triangles must be equal (4. 1.); hence AE, AC are equal. D A C B E Finally, the angle ACB being acute, as before, the adjacent angle ACD will be obtuse; since (13. 1.) these two angles are together equal to two. right angles; and the angle ADC is acute, because the angle ABD is right; consequently, the angle ACD is greater than the angle ADC; and, since the greater side is opposite to the greater angle (19. 1.); therefore the side AD is greater than the side AC. COR. 1. The perpendicular measures the true distance of a point from a line, because it is shorter than any other distance. COR. 2. Hence, also, every point in a perpendicular at the middle point of a given straight line, is equally distant from the extremities of that line. COR. 3. From the same point, three equal straight lines cannot be drawn to the same straight line; for if there could, we should have two equal oblique lines on the same side of the perpendicular, which is impossible. PROP. B. THEOR. When the hypotenuse and one side of a right angled triangle, are respectively equal to the hypotenuse and one side of another; the two right angled triangles are equal. Suppose the hypotenuse AC-DF, and the side AB-DE; the right angled triangle ABC will be equal to the right angled triangle DEF. A D Their equality would be manifest, if the third sides BC and EF were equal. If possible, suppose that those sides are not equal, and that BC is the greater. Take BH=EF (3. 1.); and join AH. The triangle ABH DEF; for the right angles B and E are equal, the side AB-DE, and BH EF; hence, these triangles are equal (4. 1.), and consequently AH=DF. Now (by hyp.), we have DF AC; and therefore, AH AC. But by the last proposition, the oblique line AC cannot be equal to the oblique line AH, which lies nearer to the perpendicular AB; therefore it is impossible that BC can differ B HC E 1 from EF; hence, then, the triangles ABC and DEF are equal. PROP. C. THEOR. Two angles are equal if their sides be parallel, each to each, and lying in the same direction. COR. If BA, AC be produced to I and H, the angle BAC=HAI; hence, the angle HAI is also equal to EDF. SCHOLIUM. The restriction of this proposition to the case where the side AB lies in the same direction with DF, and AC in the same direction with DE, is necessary; because the angle CAI would have its sides parallel to those of the angle EDF, but would not be equal to it. In that case, CAI and EDF would be together equal to two right angles. |