Therefore, it is proved, as required, that The rectangles contained by AB and AC and by AB Wherefore, If a straight line be divided, &'c. Q. E. D. 1 It may be taken as a rule in constructing the figures Props. II.-VIII. inclusive, in this Book, that the first thing to be done is to construct the square spoken of in the Enunciation, and if more than one square is mentioned, the larger is to be constructed. Exercise. If MN be a given straight line divided into any two parts in O, prove that The rectangles contained by MN and MO, and by MN and NO, together the square on MN. PROP. III. THEOREM. If a straight line be divided into any two parts, the rectangle. contained by the whole and one of the parts is equal to the rectangle contained by the two parts, together with the square on the aforesaid part. Let AB be a straight line divided into any two parts in C. CONSTRUCTION.-1. On BC describe the square CDEB (I. 46). 2. Produce ED to F, meeting AF drawn parallel to CD or BE (I. 31). PROOF.-1. It is evident that the figure AE= the sum of the figures AD and CE. of 2. But because AE is contained by AB and AF, which AF CD (I. 34) = BC (cons.), therefore AE is the rectangle AB, BC. = 3. Next, because AD is the rectangle contained by AC, AF, of which AF BC (as before), therefore AD is the rectangle AC, CB. 4. Also CE is the square on BC (cons.). Therefore, it is proved, as required, that : The rectangle AB, BC the rectangle AC, CB, together with the square on BC. Wherefore, If a straight line be divided, &c. Q. E. D. Exercise. Prove the other case of the above Proposition: The rectangle AB, AC = the rectangle AC, CB, together with the square on AC. PROP. IV. THEOREM. If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts. Let AB be a straight line, divided into any two parts in C. (I. 46) and join BD. 2. Through C draw CF parallel to AD or BE (I. 31) cutting BD in G. 3. Through G draw HGK parallel to AB or DE. PROOF.2-1. Because CF and AD are parallel, and BD falls upon them (cons.), therefore ext. angle BGC = int. opp. angle GDA (I. 29), i.e. the angle BDA (note def. 15). = Next, because AB = AD (cons.), therefore the angle ABD the angle ADB (I. 5), i.e. the angle CBG = the angle CGB (ax. 1), and therefore CB = CG (I. 6); and since CB and CG GK and BK respectively (I. 34), therefore CB, CG, GK, and KB = each other (ax. 1), and therefore the figure CK is equilateral. = 2. Further, because CG and BK are parallels, and CB falls upon them (cons.), therefore the two int. angles GCB and CBK = two right angles (I. 29). But because CBK is a right angle (cons.) therefore GCB is a right angle (ax. 1); and since the angles CBK and GCB= the angles CGK and GKB respectively (I. 34), therefore the angles CBK, GCB, CGK, and GKB each = a right angle, and therefore the figure CK is rectangular. It has also been proved to be equilateral, and therefore CK is a square (def. 30), and it is described on the side CB. 3. Similarly the figure HF is a square, on the side HG = AC (I. 34). 4. Next, because the comp. AG= the comp. GE (I. 43), and also the rect. AC, CG, the rect. AC, CB (cons.), therefore the comp. GE= the rect. AC,CB (ax. 1); and therefore the comps. AG and GE: =twice the rect. AC, CB. 5. Now, it is evident that) the whole figure ADEB But, as above, AG and GE } twice the rectangle AC, CB; Also HF and CK = the squares on AC and CB respectively; And the whole figure ADEB = the square on AB (cons.); Therefore, it is proved, as required, that The square on AB = the squares on AC and CB, together with twice the rectangle AC, CB. Cor. The parallelograms about the diameter of a square are squares. 1 In addition to the note as to Construction, Prop. II., it may now be stated that the figures of Prop. IV.-VIII. require, after the square has been described, the drawing of the diagonal, and of the parallel, or parallels, employed. 2 It will assist the learner to notice the following steps in this Proposition: a. To prove that CK is the square on CB (as in 1 and 2). b. To observe that HF is the square on HG, ¿.e. on AC (as in 3). c. To prove that the complements AG and GE twice rect. AC, CB (as in 4). = d. To combine these conclusions is the proof of the Proposition itself (as in 5). |