PROPOSITION 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 divided into any two parts at C. Then rect. AB, BC = rect. AC, BC, together with sq. on BC. A 'B Constr. On BC describe the sq. CE, (i. 46) produce ED, and through A draw AF || CD or BE. (i. 31) Dem. Then rect. AE = rects. AD, CE; but AE = rects. AB, BC, for it is contd. by AB, BE, and BE=BC; also AD = rect. AC, CB, for CD=CB; and CE is the sq. on BC; ., rect. AB, BC= rect. AC, CB together with sq. on BC. Therefore, if a straight line, &c. Corresponding result in algebra :-Let AB contain units; AC, X; BC, y units. Then b= x + y ... by = xy + y2. This should be interpreted. A little reflection will show that props. 2 and 3 are particular cases of prop. I. So in the algebraical formulæ, in the first make z=0 and a=b, and the second is the result; make >=0 and a=y, and we get the third. Thus props. 2 and 3 might have been appended to prop. i as corollaries, but Euclid considers the cases of sufficient importance to require separate and independent proofs.] PROPOSITION 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 rectangles contained by these parts. Let AB be divided into any two parts at C. Then sq. on AB=sqs. on AC and BC, together with twice rect. AC, BC. Constr. On AB describe the sq. ADEB, (i. 46) join BD, through C draw CGF | AD or BE, and through G draw HGK | AB or DE. (i. 31.) Dem. Then : BD falls on the || CF, AD, .. ext. _BGC = int. _ADB; (i. 29.) but 2 ADB=LABĎ :: BA=AD, (i. 5.) :: LCGB= L CBG; .. BC=CG; (i. 6.) but CB=GK, and CG=BK; (i. 34.) .. the figure CK is equilateral. It is also rectangular, for: CG is | BK, and CB meets them, .. Ls GCB, CBK=two rt. Ls; (i. 29.) but KBC is a rt. L ; (constr.) .. GCB is a rt. L: .. also the opp. Ls CGK, BKG are rt. Ls; (i. 34.) .. CK is rectangular, and it was proved equilateral; .. it is a sq., and it is on BC. F Similarly HF is a sq., and it is on HG, which=AC. (i. 34.) Now compl. AG=compl. GE; (i. 43.) and AG=rect. AC, BC, for GC=CB; .. AG, GE together = twice AG, that is, twice rect. AC, BC; and HF, CK are the sqs. on AC, BC; ., HF, CK, AG, GE = the sqs. on AC, BC, and twice the rect. AC, BC; but HF, CK, AG, GE make up the whole figure ABED, which is the sq. on AB; .. the sq. on AB = the sqs. on AC and BC, with twice the rect. AC, BC. Therefore, if a straight line, &c. Q. E. D. Cor. It follows from the demonstration, that the paral lelograms about the diameter of a square are also squares. [This is a prop. of great importance, and it should be thoroughly mastered before passing on. The main thing to be proved is that the figures CK, HF are squares, which might have been done more shortly by applying the corollary to prop. 46. But Euclid proves again with great clearness that if one angle of a parallelogram be a right gle, all its angles will be right angles,' and having established the point that the parallelograms about the diameter of a square are also squares, he refers. in subsequent props. (viz. 5, 6, 7) to this corollary, when the same thing has to be shown. The corresponding algebraical formula is as follows :Let AB contain a units, and let it be divided into any parts at C, so that AC contains x and BC y units. Then a = x + y ... (squaring both sides) a’ = x2 + y2 + 2xy; or, if a number be divided into any two parts, &c.] 1. From the right angle of a right-angled triangle a perpendicular is drawn on the base : show that the square on it is equal to the rectangle contained by the .. segments of the base. 2. If the square on the perpendicular from the vertex of a triangle on the base is equal to the rectangle contained by the segments of the base, the vertical angle is å right angle. 3. From one of the equal angles of an isosceles triangle a perpendicular is drawn to the opposite side : show that the square on the base is equal to twice the rectangle contained by that side and the part of it between the perpendicular and the base. 4. From one of the equal angles of an isosceles triangle a perpendicular is drawn to the opposite side :. show that the square on the perpendicular is equal to twice the rectangle contained by the segments of that side, together with the square on the line intercepted between the other equal angle and the perpendicular. PROPOSITION V. THEOREM. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Let AB be divided into two equal parts at C, and into two unequal parts at D. Then rect. AD, DB, together with sq. on CD = sq. Constr. On CB describe sq. CF, (i. 46) join BE, through D draw DHG || CE or BF, through H draw KLM || CB or EF, and through A draw AK || CL or BM. Dem. Then,:: compl. CH=compl. HF, (i. 43) add to each of these =s DM; .. the whole CM = the whole DF; but :: AC=CB, -, AL=CM, (i. 36).. also AL=DF; add to each of these equals CH, and the whole AH = CH and DF; that is, rect. AD, DB = the gnomon CMG; to each of these equals add LG, which = sq. on CD; (ii. 4. cor.) .:, rect. AD, DB together with sq. on CD = gnomon CMG together with LG: but gnomon CMG with LG make up the whole figure CF, which is sq. on CB. |