Let them meet at K. Through K draw KL || to EA or FH (I. 31). Produce HA, GB to the points L, M. Then HLKF is a gram, and HK its diameter, and AG, ME are rams about HK; and LB, BF are complements. .. LB = BF. But BFA C. .. LB = A C. And GBE = ▲ ABM (I. 15), and also Wherefore to the given straight line AB the gram LB is applied, equal to A C, and having / ABM / D.-Q.E.F. PROPOSITION XLV., PROBLEM 13. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given figure, and E the given it is required to describe a gram = ABCD, and having an angle = LE. Join DB. Describe gram FH = ^ ADB, and having ▲ FKH = ZE (I. 42). To the straight line GH apply the gram GM = DBC, and having GHM = E (I. 44). ▲ The figure FKML shall be the required ||gram. ::: ▲ E = each of 48 FKH, GHM (cons.) To each of these equals add KHG; ../s. FKH, KHG = 2s KHG, GHM (ax. 2). But s FKH, KHG =two rt. s (I. 29); ../s KHG, GHM = two rt. s, .. KHM is a straight line (I. 14). ../s HGF, HGL = two right ▲s, .. FG is in the same straight line with GL (I. 14). And because KF is || to HG, and HG to ML (cons.), KF is to ML (I. 30). And KM, FL are ||s (cons.). .. KFLM is a ||gram. And A ABD = ||gram HF, and ▲ DBC = ||gram GM; the whole figure ABCD = whole ||gram KFLM. C Wherefore the ram, &c.-Q.E.F. PROPOSITION XLVI., PROBLEM 14. To describe a square on a given straight line. Let AB be the given straight line : It is required to describe a square on AB. From point A draw AC at rt. angles to AB (I. 11). D E B Through D draw DE || to AB. For it is by construction a gram. .. AB = : ED, and AD = BE (I. 34). But AB AD. .. AB, BE, ED, DA are all equal. .. ADEB is equilateral. Again the straight line AD meets the ||s AB, DE. 28 BAD, ADE together two rt. angles. But the opposites of a gram are equal. Therefore, each of the s ABE, BED is a rt. 2. PROPOSITION XLVII., THEOREM 33. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. Let ABC be a right-angled triangle, having BAC a rt. 4. The square described on BC shall be equal to sqq. on BA, AC. H C B E Now DBC = ▲ FBA, for each of them is a rt. . Add to each ▲ ABC. DBA = whole ▲ FBC (ax. 2). FB, BD = BC, and DBA = < FBC, .. A ABDA FBC (I. 4). Now the gram BL is double of A ABD, because they are on the same base BD and between the same ||s (I. 41). And the square GB is double of ▲ FBC, ·.· they are on the same base FB, and between the same ||s (I. 41). But the doubles of equals are equal to one another (ax. 6). .. ||gram BL= square GB. In the same manner, by joining AE, BK, it can be shown that the gram CL = sq. CH. .. the whole square BDEC = two sqq. GB, HC. .. the sq. described on BC = sqq. on BA, AC. Wherefore in any right-angled triangle, &c.-Q.E.D. PROPOSITION XLVIII., THEOREM 34. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle. Let the square described on BC, a side of ▲ ABC, be equal to the squares described on the other sides BA, AC. ▲ BAC shall be a right angle. B4 From point A draw AD at right angles to AC (I. 11). Make AD = BA (I. 3). And C Join DC. Then DA = BA, sq. on DA = sq. on BA. To each of these add sq. on AC. .. The sqq. on DA, AC = sqq. on BA, AC (ax. 2). DAC is a right . The square on DC: =sqq. on DA, AC (I. 47), and by hypothesis LECTURE XVIII. BOOK II. PROPOSITIONS 1-4. In the second Book, Euclid, continuing the subject of equivalence of area, investigates the relations between the rectangles contained by straight lines divided in various ways. An oblong, or rectangle, we know is 'a figure which has all its angles right angles, but not all its sides equal' (I. def. 31), and further, since it is one of those figures classed together as parallelograms, its opposite sides are equal' (I. 34). The book opens with two definitions: 6 Def. I. Every rectangle is said to be contained by any two of the straight lines which contain one of the right angles.' With this For since the opposite sides are equal, and the angles are all right angles, these two lines being given, its area is determined, and the rectangle can be constructed. fact you are doubtless quite familiar from arithmetic, since 'any two of the straight lines which contain one of the right angles' of a rectangle will give its length and breadth, which, as you know, are the measurements required in order to find the area. Def. II. In every parallelogram any of the parallelograms about the diameter, together with the two complements, is called a gnomon.' Gnomon is a Greek word signifying a carpenter's rule. The figure is probably so called from its resemblance to a rule bent at an angle. GNO MON L |