. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. EC x EB is equal to the area of the isosceles triangle ACB;hence, QGP is the required triangle. Prop. LXVI.—In a triangle ABC let two lines be drawn from theextremities of the base BC, intersecting at any point P on the medianthrough A, and meeting the opposite sides in the points E and D;show that DE is parallel to BC. Demonstration.—Let ABC be the given triangle, P any point onthe median, and CD, BE, lines drawn fromC and B, through P. Through P, draw
. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. EC x EB is equal to the area of the isosceles triangle ACB;hence, QGP is the required triangle. Prop. LXVI.—In a triangle ABC let two lines be drawn from theextremities of the base BC, intersecting at any point P on the medianthrough A, and meeting the opposite sides in the points E and D;show that DE is parallel to BC. Demonstration.—Let ABC be the given triangle, P any point onthe median, and CD, BE, lines drawn fromC and B, through P. Through P, draw SPR parallel to BC;also, join D and E. From the similar triangles DSP andDBC, we have, DP : DC : : SP : CB . . (1). From the similar triangles EPR and EBC, we have, EP : EB : : PR, or PS : BC . . (2). From (1) and (2), we have, DP : DC : : EP : EB . . (3). Whence, by division (Bk. II, Prop. 6), we have, DP : DC — DP : : EP : EB — EP, orDP : PB : : EP : PC. Hence, the triangles DPE and BPC are similar (Bk. IV, Prop. 20),and consequently, the angles DEP and PBC are equal; hence, DE isparallel to BC, which ivas to be PROPOSITIONS FROM LEGENDRE. 49 APPLICATIONS OF ALGEBRA TO GEOMETRY. Prop. LXVII.—In a right-angled triangle ABC, given the baseBA, and the sum of the hypothenuse and perpendicular to find thehypothenuse and the perpendicular. Solution.—Denote BA by c, BC by x, AC by y, and the sum ofBC and AC by s. Then, x -\- y = s . . (1). From Bk. IV, Prop. 11, a* = y2 -f c2 . . (2). From (1), we have, x = s — y.
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