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. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB; and describe (I. 42.) the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply (I. 44.) the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. The figure FKML shall be the parallelogram required. Because the angle E is equal to each of the angles FKH, GHM, 1. The angle FKH is equal to GHM; add to each of these the angle KHG; therefore 2. The angles FKH, KHG, are equal to the angles KHG, GHM; but FKH, KHG are equal (I. 29.) to two right angles; therefore also 3. 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 the opposite sides of it, make the adjacent angles equal to two right angles, (I.14.) 4. KH is in the same straight line with HM; and because the straight line HG meets the parallels KM, FG, (I. 29.) 5. The alternate angles MHG, HGF, are equal. Add to each of these the angle HGL: therefore 6. The angles MHG, HGL, are equal to the angles HGF, HGL. But the angles MHG, HGL, are equal (I. 29.) to two right angles; wherefore also 7. The angles HGF, HGL, are equal to two right angles, and therefore (I. 14.) 8. FG is in the same straight line with GL. And because KF is parallel to HG, and HG to ML, (I. 30.) 9. and KM, FL, are parallels; wherefore (I. 34. Def.) 10. KFLM is a parallelogram; and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; 11. 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. Which was to be done. 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 (I. 44.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. 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. From the point A draw (I. 11.) AC at right angles to AB, and make (I. 3.) AD equal to AB, and through the point D draw DE parallel (I. 31.) to AB, and through B draw BE parallel to AD; therefore (I. 34. Def.) 2. AB is equal to DE, and AD to BE; but BA is equal (Constr.) to AD; therefore the four straight lines BA, AD, DE, EB, are equal to one another, and 3. The parallelogram ADEB is equilateral; likewise all its angles are right angles; for since the straight line AL meets the parallels AB, DE, (I. 29.) 4. The angles BAD, ADE, are equal to two right angles; but BAD is (Constr.) a right angle; therefore also (Ax. 3.) 5. ADE is a right angle; but the opposite angles of parallelograms are equal (I. 34.); therefore 6. Each of the opposite angles ABE, BED, is a right angle; wherefore 7. The figure ADEB is rectangular, and it has been demonstrated that it is equilateral; therefore (Def. 30.) ADEB is a square, 8. 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.-THEOREM. 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 (I. 46.) the square BDEC, and on BA, AC, the squares GB, HC; through A draw (I. 31.) AL parallel to BD, or CE, and join AD, FC. Then, because each of the angles BAC, BAG, is a right angle (Def. 30.), the two straight lines AC, AG, upon the opposite sides of AB, make with it at the point the adjacent angles equal to two right angles; therefore (I. 14.) 1. CA is in the same straight line with AG. For the same reason 2. 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, and (Ax. 2.) 3. 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 DB equal to the angle FBC; therefore (I. 4.) 4. The base AD is equal to the base FC, and the triangle ABD to the triangle FBC. Now (I. 41.) 5. The parallelogram BL is double of the triangle ABD, because they are upon the same base BD, and between the same parallels BD, AL; and 6. The square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal (Ax. 6.) to one another; therefore 7. The parallelogram BL is equal to the square GB. And, in the same manner, by joining AE, BK, it is demonstrated that 8. The parallelogram CL is equal to the square HC. Therefore (Ax. 2.) 9. 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 10. The square upon the side BC is equal to the squares upon the sides BÁ, AC. Therefore, in any right-angled triangle, &c. Q.E.D. PROP. XLVIII.-THEOREM. 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. 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 A B From the point A draw (I. 11.) AD at right angles to AC, and make AD equal to B▲, and join DC. Then because DA is equal to AB, 1. The square of DA is equal to the square of AB. To each of these add the square of AC; therefore 2. BA, AC. But (I. 47.) The squares of DA, AC, are equal to the squares of 3. The square of DC is equal 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 BĂ, AC; therefore 4. The square of DC is equal to the square of BC; and therefore also 5. 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, 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 (I. 8.) 6. The angle DAC is equal to the angle BAC. But DAC is (Constr.) a right angle; therefore also 7. BAC is a right angle. Therefore, if the square, &c. Q.E.D. |