BC is equal to the sum of the rectangles contained by the undivided line (A) and the several parts (BD, DE, and E C) of the divided line (BC). The beginner must carefully observe that the gist of the proof does not consist in showing that the rectangle BH is equal to the sum of the rectangles BK, DL, and EH, because that is self-evident; but in showing that the rectangles B H, BK, D L, and EH answer to the description of them given in the enunciation. PROPOSITION II. If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the rectangles contained by the whole and each of the parts. For the construction employed in this proposition we must be able :1. On a given straight line to describe a square. (I. 46.) 2. Through a given point to draw a straight line parallel to a given straight line. (I. 31.) Let A B be a right line divided into two parts at the point C. We have to show that the con A B is equal to the sum of rect. A B, A C and rect. A B, BC. On A B describe the A E. Through E o draw CF parallel to AD, and meet ing D E in the point F. It is self-evident that the O AE is equal to the sum of the rectangles AF and C E. But the O AE is the on A B. D A с B в A B and AC, for it is contained by AD and A C, and A D and A B are equal, being sides of the same 0. The rect. C E is the rectangle contained by A B and BC, for it is contained by BE and BC, and B E is equal to A B. Consequently the on A B is equal to the sum of the rectangles contained by the whole line (A B), and each of the parts (A C and CB) into which it is divided. Here again the beginner must observe that the main point in the above proof is, to show that the A E, and the rectangles A F and CE answer to the description of them given in the enunciation. We need no proof to show that the A E is equal to the sum of the rects. A F and C E. If in Prop. I. the two lines A and B C were of the same length, and BC were divided into two parts, it is obvious that the second proposition might be considered as only a particular case of the first ; for the rect. under A and B C would then be the same as the oon BC, and the two rects. under A and the parts of the divided line would be the same as the rectangles under the whole line BC and each of its parts. PROPOSITION III. If a right line be divided into any two parts, the rectangle contained by the whole line and one of tbe parts is equal to the square on that part together with the rectangle contained by the two parts. The propositions required for the construction em ployed in this proposition are the same as in the first and second propositions. Let A B be a right line divided into two parts at C. We have to prove that the rect. A B, A C is equal to the on A C together with the rect. AC, CB. F B А B On AC describe the O ADFC. Produce D F, and through the point B draw B E, parallel to A D, and meeting DF produced in the point E. It is self-evident that the rect. A E is equal to the sum of the O AF and the rect. CE. But A E is the rectangle contained by A B and AC, for it is contained by A B and AD, and A D and AC are equal, being sides of the same o. AF is the on A C. C E is the rectangle contained by A C and C B, for it is contained by F C and C B, and F C and A Care equal, being sides of the same o. Consequently the rectangle under the whole line (A B) and one part (A C) is equal to the square on that part, together with the rectangle contained by the two parts (A C and C B.) Here again the gist of the proof consists in showing that the rectangle A E, and its parts A F and CE answer to the description given of them in the enunciation. [The following proposition is made use of in several succeeding propositions. ] PROPOSITION A. The parallelograms about the diagonal* of a square, are themselves squares. For the proof of this proposition we must know the following, 1. Magnitudes that are equal to the same are equal to each other. * That is parallelograms whose diagonals are parts of the diagonal of the square. See Book I., Prop. 43. D F KI H fo 2. The angles at the base of an isosceles triangle are equal. (I. 5.) 3. If two angles of a triangle are equal, the sides opposite those angles are also equal. (I. 6.) 4. If a right line intersect two parallel right lines, the external angle is equal to the internal opposite angle on the same side of the intersecting line, and the two internal angles on the same side are together equal to two right angles. (I. 29.) 5. The opposite sides and angles of a parallelogram are equal. (I. 34.) Let ABCD be a 0, and A Cone of its diagonals. Through a point K G in AC the lines HG and EF are drawn parallel respectively to A B and AD. HE and F G are parallelograms about the diagonal of the square. We have to prove that they are themselves squares. First take the parallelogram HE. Because DC and HG are parallel and A C intersects them, therefore the <HKA is equal to the _DCA. But the lines A D and D C are equal because they are sides of the same 0; therefore the < DAC is equal to the < DCA. Consequently the HAK (ie., DAC) and the <HKA are equal to each other, for they are both equal to the <DCA. It follows, therefore (according to I. 6), that the line H A is equal to the line H K. But HE is a parallelogram; therefore its opposite sides are equal. Consequently H A and K E are equal, and HK and A E are equal. Therefore HE is equilateral, that is, has all its sides equal. Moreover the Zs H AE and H KE are opposite 28 A B of a parallelogram, and are therefore equal, and HAE is a right Z, therefore HKE is a right Z. Also since A B and HG are parallel and A H intersects them, the s HAE and A H K are together equal to two right Zs; and since H AE is a right Z, AHK must be a right Z, and the _KEA, which is opposite and equal to the < AHK, must also be a right Z. We have thus shown that the parallelogram HE has all its sides equal, and all its angles right angles ; therefore it is a square. In a similar way it be shown that the parallelogram F G is a square. The preceding proposition is usually incorporated in the proof of the Fourth Proposition. PROPOSITION IV. If a right line be divided into any two parts, the square on the whole line is equal to the sum of the squares on the parts together with twice the rectangle under the parts. For the construction employed in this proposition we must be able,1. To join two points by a right line. (Post. I.) 2. On a given straight line to describe a square. (I. 46.) 3. Through a given point to draw a line parallel to a given straight line. (I. 31.) For the proof of the proposition we must know,-. 1. That the parallelograms about the diagonal of a square are themselves squares. (Prop. A.) 2. That the opposite sides of a parallelogram are equal. (I. 34.) Let A B be a straight line divided into two parts at C. |