In every triangle the square on the side subtending an acute angle is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B an acute angle; and on BC, one of the sides containing it, let fall the perpendicular AD from the opposite angle. Then it is to be proved that The square on AC is less than the squares on CB and CASE I.-When the perpendicular falls within the triangle ABC. PROOF.-Because the squares on CB and DB = twice the rect. CB, DB, with the square on DC (II. 7), Therefore the squares on CB, DB, and DA = twice the rect. CB, DB, with the squares on DC and DA. But because BDA is a right angle (hyp.), Therefore the square on AB= the squares on DB and DA (I. 47). Similarly, the square on AC the squares on AD and DC. Therefore, substituting these values (in line 2), the squares on CB and AB twice the rect. CB, DB with the square on AC. Therefore, it is proved, as required in this Case, that The square on AC is less than the squares on CB and CASE II. When the perpendicular falls without the triangle ABC. B A PROOF.-Because the angle at D is a right angle (hyp.), and because the angle ACB is greater than the angle at D (I. 16), therefore the angle ACB is an obtuse angle; But because the rect. BD, BC (the rect. BC, CD, with = the square on BC (II.3), twice the rect. BC, CD, therefore twice the rect. BD, BC = with twice the square (on BC; therefore, also, substituting these values (in lines 4, &c.), the squares on AB and BC = = the square on AC, with twice the rect. BD, BC. Therefore, it is proved, as required in this Case, that The square on AC is less than the squares on CB and CASE III.—When the perpendicular is a side of the triangle ABC. B A PROOF.-In this Case the side BC is the straight line between the perpendicular let fall from the opposite angle A, and the acute angle taken at B. Now, because the square on AB = the squares on AC and CB (I. 47), therefore the squares on AB and BC= the square on AC, with twice the square on CB (ax. 2). Therefore, it is proved, as required in this Case, that The square on AC is less than the squares on CB and BA by twice the rectangle CB, CB, i.e. CB, CD, in other Cases. Wherefore, In every triangle the square, &c. Q. E. D. PROP. XIV. PROBLEM. To describe a square that shall be equal to a given Let A be the given rectilineal figure. It is required to describe a squ are the rectilineal figure A. CONSTRUCTION.-1. Describe the rectangular parallelogram BCDEA (I. 45), and if the sides BE and ED each other, it is a square, and what was required is done. = 2. But if BE and ED are not equal, then produce one of them as BE, to F, making EF = ED (post. 2, and I. 3). 3. Bisect BF in G (I. 10). 4. From centre G, with radius GB or GF, describe semicircle BHF (post. 3). 5. Produce DE to H, and join GH (posts. 2 and 1). Then it is to be proved that The square described on EH = the rectilineal figureA. PROOF.- Because the rect.EE,} EF, with the square on GE = {the square on GF (II. |