The Elements of Geometry |
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Page 7
... drawn , and but one . E D A B Let C be the given point in the straight line AB . To prove that a perpendicular can be drawn to AB at C , and but one . Draw CD , making △ BCD < △ ACD ; and let CD be re- volved about the point C as a ...
... drawn , and but one . E D A B Let C be the given point in the straight line AB . To prove that a perpendicular can be drawn to AB at C , and but one . Draw CD , making △ BCD < △ ACD ; and let CD be re- volved about the point C as a ...
Page 12
... drawn between two points . ] ( Ax . 5. ) Therefore , AE BE . - II . Let CD ( Fig . 2 ) be perpendicular to AB at its middle . point , D. Let F be any point without CD , and draw AF and BF . To prove AF > BF . Let AF intersect CD 12 ...
... drawn between two points . ] ( Ax . 5. ) Therefore , AE BE . - II . Let CD ( Fig . 2 ) be perpendicular to AB at its middle . point , D. Let F be any point without CD , and draw AF and BF . To prove AF > BF . Let AF intersect CD 12 ...
Page 14
... drawn to the line . B A Let C be the given point without the line AB , and draw CD perpendicular to AB . To prove that CD is the only perpendicular that can be drawn from C to AB . If possible , let CE be another perpendicular from C to ...
... drawn to the line . B A Let C be the given point without the line AB , and draw CD perpendicular to AB . To prove that CD is the only perpendicular that can be drawn from C to AB . If possible , let CE be another perpendicular from C to ...
Page 15
... drawn from a point to a straight line . A- -B E D Let CD be the perpendicular from C to the line AB , and CE any other line drawn from C to AB . To prove CD < CE . Produce CD to C ' , making C'D = CD , and draw EC ' . Then since ED is ...
... drawn from a point to a straight line . A- -B E D Let CD be the perpendicular from C to the line AB , and CE any other line drawn from C to AB . To prove CD < CE . Produce CD to C ' , making C'D = CD , and draw EC ' . Then since ED is ...
Page 18
... drawn from C to AB , cutting off unequal distances from the foot of the perpen- dicular , CF being the more remote . To prove CF > CE . = Produce CD to C ' , making C'D CD , and draw C'E and C'F . Then since AD is perpendicular to CC ...
... drawn from C to AB , cutting off unequal distances from the foot of the perpen- dicular , CF being the more remote . To prove CF > CE . = Produce CD to C ' , making C'D CD , and draw C'E and C'F . Then since AD is perpendicular to CC ...
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Common terms and phrases
ABC and DEF ABCD adjacent angles altitude angles are equal approach the limit arc BC area ABC bisector bisects centre chord circle circumference circumscribed cone of revolution construct the triangle Converse of Prop cylinder denote diagonals diameter diedral Draw AC equal respectively equally distant equilateral triangle equivalent exterior angle Find the area frustum given point given straight line Hence homologous hypotenuse intersection isosceles triangle lateral area lateral edges Let ABC measured by arc middle point number of sides parallelogram parallelopiped perimeter perpendicular to MN plane MN polyedral polyedrons prism produced PROPOSITION prove pyramid quadrilateral radii radius rectangle regular polygon rhombus right angles right triangle secant secant line segment similar slant height sphere spherical polygon spherical triangle square surface tangent tetraedron THEOREM trapezoid triangle ABC triangles are equal triangular prism triedral vertex vertices volume Whence
Popular passages
Page 38 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second.
Page 65 - 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.
Page 170 - 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. To prove that Proof. A Let the triangles ABC and ADE have the common angle A. A ABC -AB X AC Now and A ADE AD X AE Draw BE.
Page 120 - The first and fourth terms of a proportion are called the extremes, and the second and third terms the means.
Page 24 - Two triangles are congruent if (a) two sides and the included angle of one are equal, respectively, to two sides and the included angle of the other...
Page 123 - In any proportion the terms are in proportion by composition and division ; that is, the sum of the first two terms is to their difference as the sum of the last two terms to their difference.
Page 322 - A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The...
Page 248 - The projection of a point on a plane is the foot of the perpendicular from the point to the plane.