drawn from its vertex A, to any point Q, in BC its base: AQ+BQXQC=AB. For bisect (E. 10. 1.) BC in D, and join A, D. .. QD*+BQXQC=BD (constr. and E. 5. 2.) To each of these equals add DA2; ... AD2+QD2+BQXQC=AD*+DB': But (constr.) BD=DC, and DA is common to the ADB, ADC, and (hyp.) AB=AC; .. the 2 ADB=2 ADC; and .. each of these is a right ;. (E. 47. 1.) AD2+DQ'— AQ®, = AQ, and AD'+DB AB2; ... AQ'+BQXQC=AB2. PROP. IV. 2 5. THEOREM. The rectangle contained by the aggregate and the difference of two unequal straight lines is equal to the difference of their squares. Let AC and CB be two given unequal straight A C D B lines, of which CB is the greater; and let them be placed in the same straight line AB; so that AB is the aggregate of AC, CB, and if (E. 3. 1.) CD be cut off from CB equal AC, DB is the difference between AC and CB. Then since (constr. and E. 6. 2.) ABXDB+AC2 = CB', it is manifest, if from these equals AC' be taken, that ABXDB=CB'-AC; i. e. the rectangle contained by the aggregate AB, of AC and CB, and their difference DB, is equal to the difference of their squares. 6. COR. If there be three straight lines, the difference between the first and second of which is equal to the difference between the second and third, the rectangle contained by the first and third, is less than the square of the second, by the square of the common difference between the lines. For, let AB, CB, and DB be the three straight lines, having AC the difference of AB and CB, equal to CD, the difference of CB and DB: Then, since it has been shewn that ABXDB = CB2 AC2, it is manifest that ABXDB is less than CB' by AC'. 2 PROP. V. 7. THEOREM. The square of the excess of the greater of two given straight lines above the less, is less than the squares of the two lines, by twice the rectangle contained by them. For let AB and CB be two given straight lines, of which AB is the greater: Then is AC the excess of AB above CB; and since (E. 7. 2. ) AC*+ 2AB X BC= AB'+CB, it is manifest that AC is less than AB'+CB' by 2ABX BC. H PROP. LXXIV. 97. PROBLEM. To find a square which shall be equal to any number of given squares. First, let there be three given square, and let their sides be equal to the three straight lines A, B and C. Take any straight line DX, indefinite towards X; from D draw (E. 11. 1.) DY ❘ to DX, and produce DY indefinitely towards Y: From DX cut off (E. 3. 1.) DE A, and from DY cut off DFB; and join E, F: Again, from DY cut off DH=EF, and from DX cut off DG = C, and join G, H: The squares described (E. 46. 1.) upon GH shall be equal to the three given squares to the sides of which A, B and C are respectively equal. For (E. 47. 1. and constr.) EF2=ED2+DF2; i. e. (constr.) DH'= A2+B'; 2 .. (constr.) DH'+DG2=A2+ B2+ C2 And in the same manner, it is evident, a square may be found, which shall be equal to the aggregate of any number of given squares. PROP. LXXV. 98. PROBLEM. Two unequal squares being given, to find a third square, which shall be equal to the excess of the greater of them above the less. line, be the sides of the two given squares, of which the square of AC is the greater: From the centre C, at the distance CA, describe the circle ADE, meeting AB, produced, in E; from B draw (E. 11. 1.) BD 1 to AB, and let BD meet the circumference in D: The square of BD is equal to the excess of the square of AC above the square of BC. For join D, C: And since (constr.) the ▲ B is a right 2, .. (E. 47. 1.) CD2= CB2+BD2 i.e. (E. def. 15. 1.) AC=CB+BD Whence it is manifest, that the square of BD is equal to the excess of the square of AC above the square of CB. PROP. LXXVI. 99. THEOREM. If the side of a square be equal to the diameter of another square, the former square shall be the double of the latter. For (E. def. 30. 1. and E. 47. 1.) the square of the diameter of a square is equal to the squares of its two sides; i. e. to the double of the square itself:.. the square of any straight line which is equal to the diameter of a square, is the double of that square. PROP. LXXVII. 100. THEOREM. In any right-angled triangle, the square which is described on the side subtending the right angle, as a diameter, is equal to the squares described upon the other two sides, as diameters. For, (S. 76. 1.) the squares described on the hypotenuse, and on the two sides of a ▲ as diameters, are, respectively, the halves of the squares of those lines: But since (hyp.) the ▲ is rightangled,.. (E. 47. 1.) the square of the hypotenuse |