AB is divided, and the parts CG, GD, &c., into which CD is divided, be each of them equal to AE, it is manifest that the rectangle contained by AB and CD is equal to the square of AE taken as often as is indicated by the product of the num. ber of equal parts in AB, multiplied by the number of equal parts in CD. PROP. II. 3. THEOREM. If a straight line be divided into two unequal parts, in two different points, the rectangle contained by the two parts, which are the greatest and the least, is less than the rectangle contained by the other two parts; the squares of the two former parts, together, are greater than the squares of the two latter, taken together; and the difference between the squares of the former and the squares of the latter, is the double of the difference between the two rectangles. A Let the given straight line AB be divided into K C D B two unequal parts, in the point C, and also in the point D: Then AD X DB < ACX CB; but AD2 + DB2 > AC2 + BC2; and the excess of AD2+ DB above AC2 + CB' is the double of the excess of ACXCB above ADXDB. For, bisect (E. 10. 1.) AB in K: Therefore, ACXCB+CK* = AK' (E. 5. 2.) But CK*<DK'; ... AD×DB< ACXCB. and that, as hath been shewn ADXDB<AC×CB, ... AD2+ DB2 > AC2+CB2. Lastly, since AD*+DB'+2ADXDB =AC+CB'+2ACXCB, it is manifest, if from these equals there be taken AC2+CB'+>2ADXDB, that the excess of AD+DB above AC2+CB2 is the double of the excess of ACXCB above ADXDB. 4. THEOREM. PROP. III. In any isosceles triangle, if a straight line be drawn from the vertex to any point in the base, the square upon this line, together with the rectangle contained by the segments of the base, is equal to the square upon either of the equal sides. Let ABC be an isosceles A, and let AQ, be B Q 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. ... QD2+BQXQC=BD' (constr. and E. 5. 2.) To each of these equals add DA*; ... AD*+QD*+BQXQC=AD2+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 ;. (E. 47. 1.) AD*+DQ‘=AQ‘, and right ... AQ2+BQXQC = AB'. PROP. IV. 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 ference between AC and CB. and E. 6. 2.) ABXDB+AC2 = CB', AC, DB is the dif Then since (constr. 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=CB'AC', it is manifest that ABXDB is less than CB' by AC2. 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.) AC2+ 2AB × BC= AB'+CB', it is manifest that AC is less than AB+CB' by 2AB× BC. H PROP. VI. 8. THEOREM. The squares of any two unequal straight lines are, together, greater than twice the rectangle contained by those lines. For let AB and CB be two given straight lines of which AB is the greater: Then since (E. 7.2.) AB2+CB2 = 2AB X BC + AC2 it is manifest that AB'+ CB'>2ABX BC. PROP. VII. 9. THEOREM. If a straight line be divided into five equal parts, the square of the whole line is equal to the square of the straight line, which is made up of four of those parts, together with the square of the straight line which is made up of three of those parts. Let the straight line AB be divided into five FB C DE equal parts by the points C, D, E, F: Then, AF+AE AB2. = For since (hyp.) EF=FB.. 4FEX AF+AE= AB (E. 8. 2.) But, since AC=CD=DE= EF, 4FE=AF: |