F B H Part. Enun, and Const.-Let ABCD be a trapezium, whose side AB is || to cd. Produce co to E, and make DE AB. Through D and B draw DF and BG || to A AC (Prop. XXXI.); and through E draw Eh also Il to Ac, meeting AB produced in R. Then AHCE is a om between the same s with the trapezium ABCD, of which the base CE = CD + DE = CD + AB (Const.); and the trapezium ABCD = = $ OM AHCE. DEMONST.–For cd = AF = AB GE; .. the ACDF BGEH (Prop. XXXVI.); and the A FBD = ABDG (Prop. XXXIV.); :::, by addition, the trapezium ABCD = BDEH (Ax. 2) = į Om ACEH. Wherefore, if two sides &c.-Q. E. D. с D BF = DE DG = m m E Gen. Enun.—The sum of the areas of any two parallelograms described on the two sides of a triangle, is equal to that of a parallelogram on the base, whose side is equal and parallel to the line drawn from the vertex of the triangle to the intersection of the sides of the former parallelograms produced to meet. Part. ENUN. and Const.-Let ABC be a A, ABDE, ACFG oms described upon its sides AB, AC. Produce =ms ABDE + ACFG. Produce MB, NC to o and P. DEMONST.-1. Now, because ou is || to BA, and Bo to AH, . . ABOH is a am. I A P BMNC = K B м N 2. Also the Om ABDE = ABOH, upon the same base AB, and between the same || 8 AB, DH, (Prop. XXXV.) Om BMLK, upon base kl, and between the same || 5 HL, OB. (Prop. XXXVI.) 3. In like manner, om ACFG KLNC. 4. '.' Om ABDE + ACFG = BMLK + KLNC (Ax.2) = Om BMNC. Wherefore the sum of the areas &c.-Q. E. D. m ms PROP. XXXVII. THEOR. Gen. Enun.--Triangles upon the same base, and between the same parallels, are equal to one another. Part. Enun.—Let the A ABC, DBC be upon the same base bc, and between the same II AD, BC; then A ABC E F A DBC. B m om CONST.-Through B and c draw BE, CF respectively || to ca and bd, and meeting AD produced in E and F; .. the figures EBCA, DBCF are o ms. (Prop. XXXIV. Def.) DEMONST.-1. Now the o DBCF upon the same base BC, and between the same || $. (Prop. XXXV.) 2. But the ABC = } OM EBCA, because diam. AB bisects it; and the A DBC Ž DBCF, because diam. Dc bisects it (Prop. XXXIV.); . . A ABC = A DBC (Ax. 7). Wherefore As &c.-Q.E.D. m PROP. MM. PROB. to BF, To describe a triangle, which shall be equal to a given pentagon, and of the same altitude. Part. Enun.—Let ABCDE be the gn. pentagon. It is required to describe a A of = area, and the same altitude. Const.-Join AC, AD. Through B and E draw BF, EG respectively 11 AC and AD, and meeting cd produced in F and G. (Prop. XXXI.) Join AF, AG. Then AFG is the A required. 1. Because ac is the A ABC A AFC, upon the same base ac, and between the same || 8. (Prop. BA XXXVII.) 2. So the A AGD; .. AS ABC + AED A8 AFC + AGD (Ax. 2). 3. To each side add the A ACD; .. the pentagon ABCDE =A AFG. Wherefore a A AFG has been described pentagon ABCDE, and of the same altitude, viz. the 1 from A upon cd.-Q. E. F. AED с D to the gn. PROP. XXXVIII. THEOR. GEN. ENUN.—Triangles upon equal bases, and between the same parallels, are equal to one another. PART. ENUN. Let the A S ABC, DEF be upon = bases BC, EF, and between the same || AD; then A ABC= A DEF. Const. — Through B and r draw BG, FH respectively || to CA, ED, А H BF, B CE F. are m GBCA om and meeting ad produced in g and u (Prop. XXXI.); then the figures GBCA, DEFH oms (Def.) DEMONST.-1. Now the DEFH, upon the = base EF, and between the same || 5 GH, BF. (Prop. XXXVI.) 2. But the A ABC = į Om GBCA, because diam. AB bisects it; and the A DEF 7/ 능 OM DEFH, because diam. DF bisects it (Prop. XXXIV.); . . A ABC = A DEF (Ax. 7). Wherefore As &c. &c.—Q. E. D. Gen. Enun.- To bisect a given triangle by a line drawn from one of its angles. PART. ENUN.—Let abc be the gn. A. It is required to bisect it by a line drawn from one of its 28; for instance, the L BAC. Const.- Bisect the opposite side bc in D (Prop. X.), and join Ad. The line ad bisects the A ABC. DeMoNST.–For the A ADB = A ADC, because they are B upon bases BD, CD, and of the same altitude; i.e. between the same || $. (Prop. XXXVIII.) Wherefore the A ABC has been bisected by the st. line AD drawn from the / BAC.-Q. E. F. D Gen. Enun.- To bisect a given triangle by a line drawn from a given point in one of its sides. д D B с Part. Enun.-Let ABC be a A, D a gn. pt. in the side AC; then it is required to bisect the A ABC by a line drawn through D. Const.-Bisect sc in E (Prop. X.), and join AE, DE ; through A draw AF il to de (Prop. XXXI.), and join DF; then dF bisects the A ABC. DEMONST.-1. Because AF is || to DE, the A ADE = AFDE. (Prop. XXXVII.) 2. From each take the A GED; ..A AGD = AFGE (Ax. 3). 3. Also, because Ec = EB, the A ACE = A AEB (Prop. XXXVIII.); .., by subtraction, the trapezium dgec = trapezium ABFG (Ax. 3), and A GFE = A AGD. 4. .., by addition, the A DFC = trapezium ABFD (Ax.2). Wherefore the A ABC has been bisected by the st. line FD drawn through the gn. pt. D.-Q. E. F. D E B E GEN. Enun.-To find a point within a given triangle, from which lines drawn to the several angles will divide the triangle into three equal parts. PART. ENUN.-Let ABC be the gn. A. It is required to find a pt. within it, from which lines drawn to the A, B, C, shall divide it into three = parts. Const.–Bisect AB, BC, in D and E. (Prop. X.) Join CD, AE, intersecting in F. Then F is the pt. required. Join BF. DEMONST.-1. Because AD = DB, .. the A ADC = A BDC, and the A ADF A BDF (Prop. XXXVIII.); ..., by subtraction, A AFC = A BFC (Ax. 3). 2. Again, because BE = EC, .. the A AEB = A AEC, and the A BEF = A CEF (Prop. XXXVIII.); .., by subtraction, the A AFB = A AFC = A BFC (Ax. 1 and 3). Wherefore a pt. F has been found within the A ABC, from which the lines FA, FB, FC divide the A into three pts.-Q. E. F. |