together less than two right angles: But if a straight line fall upon two straight lines, so as to make the two interior angles on the same side of it together less than two right angles, the two straight lines, being produced, shall at length meet (Ax. 12); therefore HB, FE shall meet, if produced: Let them be produced to meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M. Then HLKF is a parallelogram, of which the diameter is HK; and AG, ME are the parallelograms about HK, and LB, BF are the complements; therefore LB is equal to BF (1. 43): But BF is equal to the triangle C; wherefore LB is equal to the triangle C: And because the angle GBE is equal to the angle ABM (1. 15), and likewise to the angle D (Constr.), therefore the angle ABM is equal to the angle D: Wherefore to the given straight line AB the parallelogram LB has been applied, equal to the triangle C, and having the angle ABM equal to the angle D. Q. E. F. PROP. XLV. PROB. 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: it is required to describe a parallelogram equal to ABCD, and having an angle equal to E. A D FGL E с KHM Join DB; and describe the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E (1. 42); and to the straight line GH apply the parallelogram GM equal to the triangle DBC, and having the angle GHM equal to the angle E (1. 44). B Then, because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM: To each of these equals add the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM: But FKH, KHG are equal to two right angles (1. 29); therefore also KHG, GHM are equal to two right angles: And because at the point H in the straight line GH, the two straight lines KH, HM, upon opposite sides of GH, make the adjacent angles together equal to two right angles, KH is in the same straight line with HM (1. 14): Again, because the straight line GH meets the parallels FG, KM, the alternate angles FGH, GHM are equal: To each of these equals add the angle HGL; therefore the angles FGH, HGL are equal to the angles LGH, GHM: But the angles LGH, GHM are equal to two right angles (1. 29); therefore also the angles FGH, HGL are equal to two right angles, and therefore FG is in the same straight line with GL (1. 14): And because FK is parallel to GH (Constr.), and GH to LM, FK is parallel to LM (1. 30) ; and FL, KM are parallels; there ore FKML is a parallelogram: And because the parallelogram FH is equal to the triangle ABD, and the parallelogram GM to the triangle DBC, the whole parallelogram FKML is equal to the whole rectilineal figure ABCD: Wherefore the parallelogram FKML has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q. E. F. COR. From this it is manifest, how to a given straight line to apply a parallelogram, equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle, viz. by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle (1. 44); &c. 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. From the point A draw AC at right angles to AB (1. 11); make AD equal to AB, and through the point D draw DE parallel to AB, and through B draw BE parallel to AD. D A Then ADEB is a parallelogram, and therefore AB is equal to DE, and AD to BE (1. 34): But AB is equal to AD; therefore the four straight lines AB, AD, DE, BE are equal to one another, and the parallelogram ADEB is equilateral: Also all its angles are right angles: For because the straight line AD falls on the two parallels AB, DE, the angles BAD, ADE are equal to two right angles (1. 29): But BAD is a right angle (Constr.); therefore also ADE is a right angle: But the opposite angles of parallelograms are equal (1. 34); therefore each of the opposite angles ABE, BED is a right angle: Therefore the figure ADEB is rectangular ; and it has been shewn to be equilateral; therefore it is a square (Def. 30), and it is described upon the given straight line AB. Q. E. F. 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 the square BDEC (1. 46), and on BA, AC the squares GB, HC; and through A draw AL parallel to BD or CE, and join AD, FC. F D L H K Then, because each of the angles BAC, BAG is a right angle, the two straight lines AC, AG, upon 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 with AG (1. 14): For the same reason, AB and AH are in the same straight line: And because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC; then the whole angle DBA is equal 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 is equal to the angle FBC-therefore the base AD is equal to the base FC, and the triangle ABD to the triangle FBC: But the parallelogram BL is double of the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL (1.41); and the square GB is double of the triangle FBC, because they are upon the same base FB, and between the same parallels FB, GC; and the doubles of equals are equal to one another; therefore the parallelogram BL is equal to the square GB: And, in like manner, by joining AE, BK, it may be demonstrated, that the parallelogram CL is equal to the square HC: Therefore the whole square BDEC is equal to the two squares GB, HC: But the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC; therefore the square upon the side BC is equal to the squares upon the sides BA, AC. Wherefore, In any right-angled triangle &c. PROP. XLVIII. THEOR. Q. E.D. If the 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. Let 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 shall be a right angle. B From the point A draw AD at right angles to AC (1. 11), and make AD equal to BA, and join DC: Then, because DA is equal to BA, the square of DA is equal to the square of BA: 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 to the squares of DA, AC, because DAC is a right angle (1. 47), and the square of BC is equal to the squares of BA, AC (Hyp.); 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 DA is equal to BA, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC, each to each, and the base DC is equal to the base BC-therefore the angle DAC is equal to the angle BAC (1. 8) : But the angle DAC is a right angle; therefore also the angle BAC is a right angle. Wherefore, If the square &c. Q. E.D. |