PROPOSITION 4.-Theorem. Let the line AB contain a linear units, and the lines AC, BC, m and n linear units respectively. Let the line AB contain 2a linear units; then its half BC will contain a Also let CD, the line between the points of section, contain m linear units. linear units. Then AD, the greater of the two unequal parts, = a + m, and DB, the less of the two unequal parts, - .. (a + m) (a — m) = a2 — m2, to each of these equals add m2, then (a + m) (a - m) + m2 = a3, = a - m, AD.DB + CD2 = BC2. PROPOSITION 6.-Theorem. Let the line AB contain 2a linear units; then its half BC will contain a linear units. Also let BD contain m linear units. Let the line AB contain a linear units, and the lines AC, BC, m and n linear units respectively. adding n2 to each of these equals, but a2 + n2 = m3 + 2mn + 2μ3, 2mn + 2n2 = 2(m + n) n = 2an, .". a2 + n2 = 2an + m2, i.e., AB+ BC2 = 2.AB.BC + AC2. PROPOSITION 8.-Theorem. Let the line AB contain a linear units, and the lines AC, BC, m and n units respectively. Let the line AB contain 2a linear units; then its half AC or BC will contain a linear units. Also let CD, the line between the points of section, contain m linear units. Let the line AB contain 2a linear units; then its half AC or BC will contain a linear units. Also let BD contain m linear units. Then the whole line and the part produced = 2a + m, and half the line and the part produced = a + m, Let the line AB contain a linear units, and AH, one of the unknown parts, x units. Let the lines BC, CA, AB contain a, b, c linear units respectively, and the lines CD, DA, m and n linear units respectively. PROPOSITION 13.-Theorem. Let the lines BC, CA, AB contain a, b, c linear units respectively, and the lines BD, AD, m and n linear units respectively. Thus 62 is less than c2 + a2 by 2a2, or 2aa, i.e., AC2 is less than AB2 + BC2 by 2.BC2 or 2.BC.BC. GEOMETRICAL EXERCISES ON BOOK II. PROPOSITION 1.-Theorem. In any triangle the squares on the two sides are together double of the squares on half the base and on the straight line joining its bisection with the opposite angle. Let ABC be a triangle, and AD the line drawn from the vertex A, to the bisection D of the base BC. Then the squares on AB, AC are double of the squares on AD, DB. Construction. From A, draw AE perpendicular to BC. Demonstration. Because ABD is an obtuse-angled triangle, therefore 1. The square on AB exceeds the squares on AD, DB, by twice the rectangle BD, DE (II. 12). And because ADC is an acute-angled triangle, therefore 2. The square on AC is less than the squares on AD, DC, by twice the rectangle CD, DE (II. 13); wherefore, since the rectangle BD, DE is equal to the rectangle CD, DE ; 3. The squares on AB, AC are double of the squares on AD, DB, Therefore, in any triangle, the squares, &c. Q.E.D. |