angle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two str. lines, and let BC be divided into any parts in D and E; then A.BC=A. BD+A. DE+A. EC Since, rect. BH = { Draw BF I BC, Prop. 10. make BG = A; Prop. 3. draw GH || BC; and DK, EL, CH || BG. Prop. 30. .. A = = BG = DK = EL. Prop. 33. Frect. BK + rect. DL + rect. EH, rect. BK = BG .BD= A.BD, A. BC = A.BD+A. DE+A.EC. 3. 2 Eu. PROP. XLII. THEOR. parts, the rectangle contained by the whole angle contained by the two parts, together with the square of the aforesaid part. Let the str. line AB be divided into two parts in the pt. C; then AB. BC = AC. BC + BC%. 30. Prop. 38. On BC descr. sq. CE. prod. ED to F. Prop. 30. draw AF || CD or BE Prop. 33 ... AF - CD BE=BC. and Def. Since, rect. AE = rect. AD + rect. CE, (rect. AE = AB. BE = AB. BC, (rect. CE = BC; Note.—If the parts of the divided line be a and b; the proposition algebraically expressed, is a+b.arab ta'. PROP. XLIII. THEOR. 4. 2 Eu. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let str. line AB be divided into any two parts in C, then ABP = AC + CB2 + 2.AC. CB. { Ax. 1. Ax. l. Upon AB descr. sq. ABED. Prop. 38. join BD CGF || AD or BE, BA= AD, Prop. 5. _BGC, Prop. 6. but BC = GK, Prop. 33. .. CG = GK; fig. CK is equilat. Again :: CB meets Ils CG, BK, .. LKBC + ZGCB = 2 rt. Zs; Prop. 28. but LKBC < GCB = rt. L Prop. 33. . . fig. CK is rectang. fig. CK = CB. In the same way it may be proved, that fig HF = AC-. Also fig. AG = fig. GE; Prop. 37. i. e. fig. AG= AC.CG -- AC.CB. or rt. L Ax.l. fig. GE= AC.CB AG, GE, i. e. the SAC + CB + 2. AC. CB. Cor. The about the dia. of a sq. are likewise squares. Note. If the parts of the divided line be a and b, the proposition algebraically expressed is (a+b)'=a + 2ab +62. yms PROP. XLIV. THEOR. 5. 2 Eu. 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 of the line between the points of section, is equal to the square of half the whole line. Let the str. line AB be divided into two equal parts at the pt. C, and into two unequal parts at the point D; then AD. DB + CD = CB2. . fig. CM ={fig. AL, for AC= CB; On CB descr. sq. CEFB, Prop. 38. join BE; ŞAK, DHG || CE or BF Prop 30. draw KLHM || AB. add to each the fig. DM. Prop. 35. .. fig. AL = fig. DF, add to each the fig. CH, :: whole fig. AH = fig. DF+fig. CH, but fig. AH = JAD. DB, for DH Çor. Prop. 43. = DB; :: gnomon CMG = AD. DB, Prop. 43. :: gnomon CMG+fig. LG = AD.DB+CD2, i. e. whole fig. CEFB or CB’=AD.DB+CD2. Wherefore, if a str. line, &c. Cor.—Since CB2 – CD=AD.DB; the diff. of the Squares of two unequal lines is equal to the rectangle of their sum and diff. Note.--If the unequal parts of the divided line be b and c, and the half line be a; the proposition algebraically expressed is bc + (a - b)=a”; or bc + (c-a) = a. Cor. 6. 2 Eu. PROP. XLV. THEOR. E |