. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. ROM LEGENDRE. 47 But, from the same proposition, we have, CF2 + AF2 = 2 AE2 -f 2EF2;which gives, when substituted in the preceding equation, CD2 + CB2 + AD2 + AB2 = 4BF2 + 4AE2 + 4EF2. But, 4BF2 = BD2, (Bk. IV, Prop. 8, Cor.); also 4AE2 = AC25hence, CD2 + CB2 _j_ AD2 + AB2 = BD2 + AC2 + 4EF2,which was to be proved. Prop. LXV.— Construct an equilateral triangle equal in area to any isosceles triangle. Solution.—Let ABC be any isosceles triangle. On AB, a
. Selected propositions in geometrical constructions and applications of algebra to geometry. Being a key to the appendix of Davies' Legendre. ROM LEGENDRE. 47 But, from the same proposition, we have, CF2 + AF2 = 2 AE2 -f 2EF2;which gives, when substituted in the preceding equation, CD2 + CB2 + AD2 + AB2 = 4BF2 + 4AE2 + 4EF2. But, 4BF2 = BD2, (Bk. IV, Prop. 8, Cor.); also 4AE2 = AC25hence, CD2 + CB2 _j_ AD2 + AB2 = BD2 + AC2 + 4EF2,which was to be proved. Prop. LXV.— Construct an equilateral triangle equal in area to any isosceles triangle. Solution.—Let ABC be any isosceles triangle. On AB, as aside, construct the equilateral triangleABD; draw the line DCE, which is thecommon bisectrix of the angles D and C. On DE, as a diameter, draw a semi-circle DFE, and from C erect the line CFperpendicular to DE; draw EF; andfrom E, as a centre, with EF as a radius,draw the arc FG-, cutting DE in G;through G, draw GP and GQ, parallel to DB and DA. Also draw FD. From Prop. 23, Bk. IV, Cor. 2, we have, EF2, or EG2 = EC x ED, orED : EG : : EG : EC . . (1).From the similar triangles EDB and EGP, we have,ED : EG : : EB : EP . . (2).. 48 KEY. From (1) and (2), we have, EG : EC : : EB : EP, or EG x EP = EC x EB; but, EG x EP is equal to the area of the equilateral triangle QGP,and 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 .
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