Schultze and Sevenoak's Plane and Solid Geometry |
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Page 11
... a radius sufficiently large draw two arcs intersecting in F Then the line BF is the required bisector . III . At a point C in a given straight line AB , to draw an angle equal to a given angle O. E Да FB From O as a center , with any ...
... a radius sufficiently large draw two arcs intersecting in F Then the line BF is the required bisector . III . At a point C in a given straight line AB , to draw an angle equal to a given angle O. E Да FB From O as a center , with any ...
Page 12
... given point C in AB draw a perpendicular to AB . Divide a given angle into 4 equal parts . Ex . 49 . Divide a given angle into 8 equal parts . Ex . 50 . Construct an angle of ( a ) 90 ° ( b ) 45 ° . Ex . 51 . Construct an angle of ( a ) ...
... given point C in AB draw a perpendicular to AB . Divide a given angle into 4 equal parts . Ex . 49 . Divide a given angle into 8 equal parts . Ex . 50 . Construct an angle of ( a ) 90 ° ( b ) 45 ° . Ex . 51 . Construct an angle of ( a ) ...
Page 17
... given point in a given line only one perpen- dicular can be drawn to the line . For if two perpendiculars , AD and AC , could be drawn at A , we should have two unequal right angles BAD and BAC , which is impossible ( 46 ) . D C A 48 ...
... given point in a given line only one perpen- dicular can be drawn to the line . For if two perpendiculars , AD and AC , could be drawn at A , we should have two unequal right angles BAD and BAC , which is impossible ( 46 ) . D C A 48 ...
Page 25
... Given ABC and A'B'C ' with AB A'B ' , BC = B'C ' , and LB = LB ' . To prove AABC = A A'B'C ' . STATEMENTS Proof REASONS Place A ABC upon △ A'B'C ' so that BC shall coincide with B'C ' . BA will take the direction of B'A ' , The point 4 ...
... Given ABC and A'B'C ' with AB A'B ' , BC = B'C ' , and LB = LB ' . To prove AABC = A A'B'C ' . STATEMENTS Proof REASONS Place A ABC upon △ A'B'C ' so that BC shall coincide with B'C ' . BA will take the direction of B'A ' , The point 4 ...
Page 39
... a given angle . Ex . 184. Construct the angle - bisectors of a given triangle . [ For prac- tical applications see page 286. ] PROPOSITION IX . PROBLEM 85. At a given point in a given straight line , to erect a perpendicular to that line .
... a given angle . Ex . 184. Construct the angle - bisectors of a given triangle . [ For prac- tical applications see page 286. ] PROPOSITION IX . PROBLEM 85. At a given point in a given straight line , to erect a perpendicular to that line .
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Common terms and phrases
altitude angle equal angle formed angles are equal annexed diagram bisect bisector chord circumference circumscribed congruent cylinder diagonals diagram for Prop diameter dihedral angles divide draw drawn equiangular polygon equilateral triangle equivalent exterior angle face angles Find the area Find the radius Find the volume frustum given circle given line given point given triangle Hence HINT homologous hypotenuse inscribed intersecting isosceles triangle lateral area lateral edge line joining locus median parallel lines parallelogram parallelopiped perimeter perpendicular plane MN polyhedral angle polyhedron prism PROPOSITION prove Proof pyramid Q. E. D. Ex quadrilateral radii ratio rectangle reflex angle regular polygon respectively equal rhombus right angles right triangle segments sphere spherical polygon spherical triangle square straight line surface tangent THEOREM trapezoid triangle ABC triangle are equal trihedral vertex angle vertices
Popular passages
Page 222 - Two triangles having 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.
Page 208 - The area of a rectangle is equal to the product of its base and altitude.
Page 183 - If two chords intersect in a circle, the product of the segments of one is equal to the product of the segments of the other.
Page 160 - A line parallel to one side of a triangle divides the other two sides proportionally.
Page 82 - 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 70 - If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first triangle 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 188 - Pythagorean theorem, which states that the sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse.
Page 409 - A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the sides of the polygon ; the...
Page 333 - The sum of any two face angles of a trihedral angle is greater than the third face angle.
Page 193 - In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it.