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ABCD according adjacent alternate altitude antecedents apply base bisects called chord circumscribed coincide common Compare compounded consequents construct contained converse corresponding Definition describe diagonal difference divided double Draw drawn equal equilateral equivalent extremities figure four geometry given angle given circle given line given point greater half Hence hypothesis inscribed interior angles intersect isosceles less limit line joining locus magnitudes mean measure meet multiple Note number of sides opposite sides Outline pair parallel parallelogram passes perimeter perpendicular placed polygon portion position PROBLEM prolonged proof proportional prove quadrangle radii radius ratio rectangle regard regular polygons relation remaining right angle scale segments Show sides similar similarly solution square statement straight angle straight line successive surface symmetric Take taken tangent THEOREM third triangle triangle ABC true unequal vertex whole
Page 148 - In every triangle, the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.
Page 7 - LET it be granted that a straight line may be drawn from any one point to any other point.
Page 136 - If a straight line be divided into any two parts, the square on the whole line is...
Page 195 - UPON a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle.
Page 80 - The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it 46 INTERCEPTS BY PARALLEL LINES.
Page 305 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Page 287 - ... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
Page 263 - If four magnitudes of the same kind be proportionals, they shall also be proportionals when taken alternately. Let A, B, C, D be four magnitudes of the same kind, which are proportionals, viz.