Book II. BGC is equal to the interior and opposite angle ADB ; but ADB is equals to the angle ABD, because BA is equal 29. 1. to AD, being, sides of a square; Do lo wherefore the angle CGB is equal to the angle GBC; and therefore and CG to BK; wherefore the F E therefore equal to two right angles; and KBC is a right angle; wherefore GCB is a right angle: and therefore also the angles CGK, GKB opposite to these, are right angles, and CGKB is rectangular; but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB: For the same reason HF also is a square, and it is upon the side HG, which is equal to AC: Therefore HF, CK are the squares of AC, CB; and because the comple$ 43. 1. ment AG is equals to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB: And HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB : But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB : Therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore if a straight line, &c. Q.E.D. Cor. From the demonstration, it is manifest, that the parallelograms about the diameter of a square are likewise squares. Boox II. PROP. V. THEOR. If a straight line be divided into two equal parts, Upon CB describea the square CEFB, join BE, and · 46. 1. through D drawb DHG parallel to CE or BF; and through 31. 1. H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM: And because the complement CH is equal to the complement HF, to each of < 43. 1. these add DM; therefore the whole CM is equal to the whole DF; but CM is equald A D B 436. 1. to AL; because AC is equal to CB; therefore also ALis equal to DF. To each of K these add CH, and the whole AH is equal to DF and CH: But AH is the rectangle contained by AD, DB, for DH is equale to DB; and DF together with CH Cor. 4. 2. is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB: To each of these add LG, which is equale to the square of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD; But the gnomon CMG. and LG make up the whole figure CEFB, which is the square of CB: Therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore if a straight line, &c. Q. E. D. From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference. L 146. 1. Upenf CB, is equa rectangle A Book II. PROP. VI. THEOR. If a straight line be bisected, and produced to any point: the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal 'to the square of the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D; the rectangle AD, DB, together with the square of CB, is equal to the square of CD. a 46. 1. Upon CD describe a the square CEFD, join DE, and 131. 1. through B Grawb BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through B D therefore also AL is equal F ( F. • Cor. 4. 2 is the rectangle contained by AD, DB, for DM is equale to DB: Therefore the gnomon CMG is equal to the rectangle AD, DB: Add to each of these LG, which is equal to the square of CB; therefore the rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG, and the figure LG; But the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D. PROP. VII. THEOR. If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC are equal to twice the Book II. rectangle AB, BC, together with the square of AC. Upon AB describe the square ADEB, and construct the 146. 1. figure as in the preceding propositions; and because AG is equalb to GE, add to each of them CK; the whole AK is o 43. 1. therefore equal to the whole CE; therefore AK, CE, are double of A AK: But AK, CE, are the gnomon AKF, together with the square CK; therefore the gnomon AKF, toge-H ther with the square CK, is double of AK: But twice the rectangle AB, BC is double of AK, for BK is equale to BC: Therefore the gno Cor. 4. .. mon AKF, together with the square D CK, is equal to twice the rectangle AB, BC: To each of these equals add HF, which is equal to the square of AC; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC : but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC: therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC. Wherefore, if a straight line, &c. Q. E. D., PROP. VIII. THEOR. If a straight line be divided into any two parts; four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line, which is made up of the whole and that part. Let the straight line AB be divided into any two parts in the point C; four times the rectangle AB, BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together. Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equala to GK, and BD to KN; therefore GK is a 34. 1. ang so are 9 BD, what is, dis, Book IĮ. equal to KN: For the same reason, PR is equal to RO; and m because CB is equal to BD, and GK to KN, the rectangle b36. 1. CK is equalb to BN, and GR to RN; but CK is equalc to €43. 1. RN, because they are the complements of the parallelogram CO; therefore also BN is equal to GR; and the four rectangles BN,CK, GR, RN are therefore equal to one another, and so are quadruple of one of them CK: Again, because CB is equal to BD, and that BD is . Cor. 4. 2. equald to BK, that is, to CG, and CB equal to GK, that d is, to AG is equal to MP, and PL to, € 43. 1. RF: But MP is equale to PL, A because they are the complements E H L the rectangle AB, BC, is equal to the gnomon AOH. To Cor. 4. 2. each of these add XH, which is equalf to the square of AC: Therefore four times the rectangle AB, BC together with the square of AC, is equal to the gromon AOH and the square XH: But the gnomon AOH and XH make up the figure AEFD, which is the square of AD: Therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D. . |