And because at the point H in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles, KH is in the same straight line with HM. [I. 14. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal. [I. 29. Add to each of these equals the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL. [Axiom 2. But MHG, HGL are together equal to two right angles; [I. 29. therefore HGF, HGL are together equal to two right angles Therefore FG is in the same straight line with GL. [I. 14. And because KFis parallel to HG, and HG to ML,[Constr. KF is parallel to ML; and KM, FL are parallels ; therefore KFLM is a parallelogram. [I. 30. [Construction. [Definition. And because the triangle ABD is equal to the parallelogram HF, [Construction. and the triangle DBC to the parallelogram GM; [Constr. the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. [Axiom 2. Wherefore, the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q.E.F. COROLLARY. 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; namely, by applying to the given straight line a parallelogram equal to the first triangle ABĎ, and having an angle equal to the given angle; and so on. [I. 44. PROPOSITION 46. PROBLEM. To describe a square on a given straight line. Let AB be the given straight line: it is required to describe a square on AB. From the point A draw AC at right angles to AB; [I. 11. and make AD equal to AB ; [I. 3. ADEB shall be a square. For ADEB is by construction a parallelogram ; therefore AB is equal to DE, and AD to BE. [I. 34. But AB is equal to AD. D [Construction. Therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral. Likewise all its angles are right angles. [Axiom 1. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles; but BAD is a right angle ; therefore also ADE is a right angle. [I. 29. [Construction. [Axiom 3. But the opposite angles of parallelograms are equal. [I. 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. [Axiom 1. [Definition 30. And it is described on the given straight line AB. Q.E.F. COROLLARY. From the demonstration it is manifest that every parallelogram which has one right angle has all its angles right angles. PROPOSITION 47. THEOREM. In any right-angled triangle, the square which is de scribed on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. Let ABC be a right-angled triangle, having the right angle BAC: the square described on the side BC shall be equal to the squares described on the sides BA, AC. On BC describe the square BDEC, and on BA, AC de scribe the GB, HC; squares [I. 46. H the two straight lines AC, AG, on 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 with AG. [I. 14. For the same reason, AB and AH are in the same straight line. Now the angle DBC is equal to the angle FBA, for each of them is a right angle. Add to each the angle ABC. [Axiom 11. [Axiom 2. Therefore the whole angle DBA is equal to the whole angle FBC. And because the two sides AB, BD are equal to the two sides FB, BC, each to each; and the angle DBA is equal to the angle FBC; therefore the triangle ABD is equal to the triangle FBC. [Definition 30. [I. 4. Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL. [I. 41. And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels FB, GC. [I. 41. But the doubles of equals are equal to one another. [Ax. 6. Therefore the parallelogram BL is equal to the square GB. In the same manner, by joining AE, BK, it can be shewn, that the parallelogram ČZ is equal to the square CH. Therefore the whole square BDEC is equal to the two squares GB, HC. [Axiom 2. And the square BDEC is described on BC, and the squares GB, HC on BA, AC. Therefore the square described on the side BC is equal to the squares described on the sides BA, AC. Wherefore, in any right-angled triangle &c. Q.E.D. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle. Let the square described on BC, one of the sides of the triangle ABC, be equal to the squares described on the other sides BÁ, AC: the angle BAC shall be a right angle. From the point A draw AD at right angles to AC; [I. 11. and make AD equal to BA; '[I. 3. on AC. B D Therefore the squares on DA, AC are equal to the squares on BA, AC. [Axiom 2. But because the angle DAC is a right angle, [Construction. the square on DC is equal to the squares on DA, AC. [I. 47. And, by hypothesis, the square on BC is equal to the squares on BA, AC. Therefore the square on DC is equal to the square on BC.[Ax.1. Therefore also the side DC is equal to the side BC. And because the side DA is equal to the side AB; [Constr. and the side AC is common to the two triangles DAC, BAC; the two sides DA, AC are equal to the two sides BA, AC, each to each; and the base DC has been shewn to B be equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8. 1. EVERY right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles. 2. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. |