Inhaltsangabe
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. 1893 edition. Excerpt: ... N, the point of intersection, is the center, and N ft = NG; and,..., drawing RSOK, GT=TS (B. II., P. xxi., C. i.), and, also, TS = BS in Abto; that is, ffG is trisected, and T is the centroid or intersection of the medians (B. II., P. xxv., C).-Then, the centroid, the center of the nine-points ©, the orthocenter and the center of the circumscribed 0 s are collinear. BOOK VI.--MENSURATION: PLANE FIGURES. SECTION II.--TRIANGLES. 349. i. Bisect a given triangle by a line drawn from one of its angles. Draw the line to the middle of the opposite side. (P. VI.) 2. Find a point within a given triangle from which lines drawn to the several angles-will divide the triangle into three equal parts. I. The intersection of the medians; as-sume construction and demonstration of B. II., P. xxv. and C. 2. A B H C 02 A HF Co Aahc (P. vi.); and, Abhco2aehboabh (P. VI.); that is, A AB H oBHC oAHC. 3. Trisect a given triangle from a given point within it. I. Trisect base B C, points of division being E and F. Draw 0 E and OF from given point 0. Draw AGOE and AHOF. Draw AE and AF; also, OA.OG and OH. 2. 0A, OG, and OH are the re-quired lines. A E cuts off J of A ABC (P. vi.); i.e., A ABE O i A A B C; but, A 0 G B o A ABE; for, A Goeoaaoe (B. III., P. XX., C. II.), being on same base OE and between s A G and 0 E. Take away common part A K 0 E, and, then, A GKEo Aako;.:, Aabe-agke+ AAKOoi A ABC; but, A ABE-A GKE + A AKO is figure AOGB; i.e.,AOGBo$AABC. Similarly, it is proved that AOHC Ok A ABC; and,.-., Aogho Aabc. 4. Construct a triangle, the base, an adjacent angle, and the sum of the two other sides being given. (See § I4I, Ex. 30.) 5. Construct a triangle, the base, an adjacent angle, and the difference of the two other sides being given. (See §...
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