This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1883 Excerpt: ...AB'BH = MBBL, but BlI = BC, and BL = BD; . ABBC 3IBBD. EXERCISE 188. If an arc of a circle be cut equally and unequally, prove that tie rectangle under the chords of the unequal segments, together with the square of the chord of the intermediate arc, is equal to the square of the chord of half the arc. (Fig. 188, Plate ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1883 Excerpt: ...AB'BH = MBBL, but BlI = BC, and BL = BD; . ABBC 3IBBD. EXERCISE 188. If an arc of a circle be cut equally and unequally, prove that tie rectangle under the chords of the unequal segments, together with the square of the chord of the intermediate arc, is equal to the square of the chord of half the arc. (Fig. 188, Plate XIII.)--Let the arc AEB of the circle AM BE be bisected in E, and divided unequally in F; then the rectangle AF'FB together with EF2 will be equal to AE2. Join AB, AE, EF, EB, FB, AF; draw the diameter EM, cutting AB at C, and draw FD; and from /'draw FH parallel to AB, and cutting EM in N. Because the arc AEB is bisected in E, and its chord AB bisected perpendicularly by EM, then EM passes through the centre of the circle (III. 1); and since FH is parallel to AB it is also bisected in N. Now, in the triangle AFB, since FD is perpendicular to AB, the rectangle AFFB = ME-FD Ex. 187); for the same reason AE-EB = AE2 = 3lE-EC = MEEN + ME-NC = ME-EN + ME-FD. But MM EN = HEEF = EF2; and.'. AE2 = AFFB + EF2 (Ax. 1). EXERCISE 189. Given the base, vertical angle, and the sum of the squares of the sides of a triangle: to construct it. (Fig. 189, Plate XIII.)--Let AB be the given base. On it describe a segment of a circle containing an angle equal to the given vertical angle (III. 33); bisect AB in C; find a line equal to half the difference between the sum of the squares of the sides and twice the square of half AB (Ex. ); from C as centre, with this line as distance, describe an arc cutting the circumference of the segment in D; join AD and DB; then ADB is the triangle required. Because AB is bisected in C, and CD joined, AD2 + DB2--2AC2 + 2CD2 (Ex. 255, Part I.). EXERCISE 190. Given the base, vertical angle, and the difference of the squares of the ...
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