draw cL parallel to Ad Or BE (I. 27.); and join BF, CD, AK and ce. Then, since the right line ac meets the two right lines GC, CB in the point c, and makes each of the angles ACG, ACB a right angle (by Hyp. and Def. 2.), GC will be in the same right line with CB (I. 14.) And, because the angle Fac is equal to the angle DAB (I. 8.), if the angle CAB be added to each of them, the whole angle FAB will be equal to the whole angle DAC. The sides fa, AB, are, also, equal to the sides ca, AD, each to each, (Def. 2.), and their included angles have, likewise, been shewn to be equal; whence the triangle ABF is equal to the triangle ACD (I. 4.) But the square ag is double the triangle ABF (I. 32.), and the parallelogram Al is double the triangle ACD (1 32.); consequently the parallelogram Al is equal to the {quare AG (Ax. 6.) And, in the same manner, it may be demonstrated, that the parallelogram bl is equal to the square BH; therefore the whole square ae is equal to the squares AG and bh taken together. Q. E, D. COROLL. The difference of the squares of the hypotenuse and either of the other fides is equal to the square of the remaining fide. A PRO P. XV. THEOREM. If the square of one of the sides of a triangle be equal to the sum of the squares of the other two sides, the angle contained by those fides will be a'right angle. A Let ABC be a triangle ; then if the square of the side AB be equal to the sum of the squares of ac, CB, the angle ACB will be a right angle. For, at the point c, make cd at right angles to CB (I. 11.), and equal to Ac (I. 3.); and join DB. Then, since the squares of equal lines are equal (II. 2.), the square of Dc will be equal to the square of Ac. And, if, to each of these equals, there be added the square of CB, the squares of DC, CB will be equal to the squares of AC, CB. But the squares of DC, CB are equal to the square of BD (II. 14.), and the squares of AC, CB to the square of AB (by Hyp.); whence the square of BD is equal to the square of AB. And since equal squares have equal fides (II. 3.), AB is equal to BD; BC is also common to each of the tri. angles ABC, DBC, and ac is equal to cd (by Confl.); con consequently the angle ACB is equal to the angle BCD (I. 7.) But the angle bed is a right angle (by Conf.), whence the angle ACB is also a right angle. Q. E, D. PRO P. XVI. THEOREM, The difference of the squares of the two sides of any triangle, is equal to the difference of the squares of the two lines, or distances, included between the extremes of the base and the perpendicular. Let ABC be a triangle, having cd perpendicular to AB; then will the difference of the squares of ac, co be equal to the difference of the squares of AD, DB. For the sum of the squares of AD, DC is equal to the square of Ac (II. 14.); and the sum of the squares of BD, DC is equal to the square of BC (II. 14.) The difference, therefore, between the sum of the squares of AD, dc and the sum of the squares of BD, DC, is equal to the difference of the squares of AC, CB. And, since dc is common, the difference between the fum of the squares of AD, DC, and the sum of the squares of BD, DC is equal to the difference of the squares of AD, DB. But things which are equal to the same thing are equal to each other ; consequently the difference of the squares of AC, CB is equal to the difference of the squares of AD, DB. Q. E. D. COROLL. The rectangle under the sum and difference of the two sides of any triangle, is equal to the rectangle under the base and the difference of the segments of the base (II. 13.) In any obtuse-angled triangle, the square of the side subtending the obtuse angle, is greater than the sum of the squares of the other two sides, by twice the rectangle of the base and the distance of the perpendicular from the obtuse angle, B Let ABC be a triangle, of which ABC is an obtufe angle, and CD perpendicular to AB; then will the square of AC be greater than the squares of AB, BC, by twice the rectangle of AB, BD. For, since the right line Ad is divided into two parts, in the point B, the square of ad is equal to the squares of F AR, AB, BD, together with twice the rectangle of AB, BD (II. 11.) And if, to each of these equals, there be added the fquare of DC, the squares of AD, DC will be equal to the squares of AB, BD and Dc, together with twice the rectangle of AB, BD. But the squares of AD, DC are equal to the square of AC, and the squares of BD, DC to the square of BC (II. 14.); whence the square of Ac is greater than the squares of AB, BC by twice the rectangle of AB, BD. Q. E. D. PRO P. XVIII. THE OR EM. In any triangle, the square of the side suba tending an acute angle, is less than the sum of the squares of the base and the other side, by twice the rectangle of the base and the distance of the perpendicular from the acute angle. AN B. Let ABC bé a triangle, of which ABC is an acute an. gle, and cd perpendicular to AB: then will the square of AC, be less than the sum of the squares of AB and BC, by twice the rectangle of AB, BD. For, since AB, and AB produced, are divided into two parts in the points D, and A, the sum of the squares of AB, |