DE parallel to AB (I. 31); and through B, draw BE parallel to AD, meeting DE in E. wherefore D A Then ABED is a square. Proof. Because DE is parallel to AB, and BE parallel to AD (constr.), therefore 1. ABED is a parallelogram; wherefore B 2. AB is equal to DE, and AD to BE (I. 34); but BA is equal to AD (constr.); therefore the four straight lines AB, BE, ED, DA are equal to one another, and 3. The parallelogram ABED is equilateral. It has likewise all its angles right angles, since AD meets the parallels AB, DE, therefore 4. The angles BAD, ADE are equal to two right angles (I. 29), but BAD is a right angle (constr.); therefore also 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; 7. The figure ABED is rectangular, and it has been proved to be equilateral; therefore 8. The figure ABED 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. PROPOSITION 47.-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. F Let ABC be a right-angled triangle, having the right angle BAC. Then the square described upon the side BC shall be equal to the squares described upon BA, AC. Construction. On BC describe the square BDEC (I. 46); and on BA, AC, the squares GB, HC; through A draw AL parallel to BD or CE (I. 31), and join AD, FC. Demonstration. Then because the angle BAC is a right angle (hyp.), and that the angle 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 A, the adjacent angles equal to two right angles; therefore 1. CA is in the same straight line with AG (I. 14). For the same reason, 2. BA 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 of these equals the angle ABC, therefore 3. The whole angle ABD is equal to the whole angle FBC (Ax. 2). And because the two sides AB, BD are equal to the two sides FB, BC, each to each, and the included angle ABD is equal to the included angle FBC, therefore Now 4. The base AD is equal to the base FC (I. 4); and the triangle ABD to the triangle FBC. 5. The parallelogram BL is double of the triangle ABD (L. 41), because they are upon the same base BD, and between the same parallels BĎ, AL; also 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 to one another (Ax. 6); therefore 7. The parallelogram BL is equal to the square GB. Similarly, by joining AE, BK, it can be proved that 8. The parallelogram CL is equal to the square HC. Therefore 9. The whole square BDEC is equal to the two squares GB, HC (Ax. 2); and the square BDEC is described upon the straight line BC, and the squares GB, HC upon AB, AC; therefore 10. The square upon the side BC is equal to the squares upon the sides AB, AC. Therefore, in any right-angled triangle, &c. Q.E.D. PROPOSITION 48.-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. Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other two sides, AB, AC. B Then the angle BAC shall be a right angle. Construction. From the point A draw AD at right angles to AC (I. 11); make AD equal to ÃB, and join B DC Demonstration. Then, because AD is equal to AB, 1. The square on AD is equal to the square on AB, to each of these equals add the square on AC; therefore but 2. The squares on AD, AC are equal to the squares on AB, AC; 3. The squares on AD, AC are equal to the square on DC (I. 47), because the angle DAC is a right angle; and the square on BC, by hypothesis, is equal to the squares on BA, AC; therefore 4. The square on DC is equal to the square on BC; and therefore 5. The side DC is equal to the side BC. And because the side AD is equal to the side AB, and AC is common to the two triangles DAC, BAC; the two sides DA, AC are equal to the two BA, AC, each to each; and the base DC has been proved to be equal to the base BC; therefore 6. The angle DAC is equal to the angle BAC (I. 8) ; but DAC is a right angle; therefore also 7. BAC is a right angle. Therefore, if the square described upon, &c. Q.E.D. BOOK II. 1. EVERY right-angled parallelogram is called a rectangle, and is said to be contained by any two of the straight lines which contain one of the right angles. A DEFINITIONS. 2. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a gnomon. A E K "Thus the parallelogram HG, together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon." PROPOSITION 1.-Theorem. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. |