The Art of Drawing in Perspective, from Mathematical Principles

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. 1805. Excerpt: ... PROPOSITION I. PROBLEM. 20 describe an equilateral triangle, upon a given right line. Let AB be the given right line, upon which it is required to describe an equilateral triangle. About the centre A, with the distance AB, describe the circle BCD a; and about the centre B, with the distance BA, describe, Postuiatc the circle ACE % from the point C, where the two circles inter-3. sect each other, draw the lines CA, CB b. b post. x Then, because A is the centre of the circle DBC; AC is equal to AB c; and because B is the centre of the circle ACE; c Definition BC is equal to BA c; but CA is proved equal to AB; and BC 15. to AB; therefore BC is equal to AC d; and the three sides AB, d Axiom 1. BC, CA, are equal to one another: therefore, upon the given right line AB, there is described an equilateral triangle ABC: which was required. PROP. II. PROB. From a given point to draw a right line equal to a given right line. Let A be the given point, and BC the given right line, it is required from the point A, to draw a right line, equal to the given right line BC. With the centre C, and distance BC, describe the circle BGH a; a Post join AC b, upon which describe an equilateral triangle DAG c;bPost.'i.' produce DC passing through the centre, to G, in the circumfer-c 1ence; and DA to any distance E d, not less than DG; withdPst-athe centre D, and distance DG, describe the circle KGL. Then, because Cis the centre of the circle BGH; BC is equal to CG e; and because D is the centre of the circle KGL; DGe Def.15. is equal to DL e; but DC is equal to DA c; therefore the remainders AL, and CG, are equal f; but BC, AL, are each f Ax. 3. proved equal to CG; and therefore equal to one another -: g i, . therefore, from the point A, there is drawn the right line AL equal t..