The Principles of the Solution of the Senate-house 'riders,' Exemplified by the Solution of Those Proposed in the Earlier Parts of the Examinations of the Years 1848-1851 |
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Page 1
... opposite sides and angles of parallelograms are equal to one another ; and the diameter bisects them . ( Euc . I. 34. ) ( B ) . If the opposite sides or the opposite angles of any quadrilateral figure be equal , or if its diagonals ...
... opposite sides and angles of parallelograms are equal to one another ; and the diameter bisects them . ( Euc . I. 34. ) ( B ) . If the opposite sides or the opposite angles of any quadrilateral figure be equal , or if its diagonals ...
Page 3
... opposite sides of it : join CD meeting AB in E ; shew that CE is equal to ED . 0 Make ABD ' = 2ABD and BD ' = BD ( fig . 3 ) . Join CD ' , AD ' , ED ' . By construction , AABD ' = ABD , ABC , by hypothesis . By applying the proposition ...
... opposite sides of it : join CD meeting AB in E ; shew that CE is equal to ED . 0 Make ABD ' = 2ABD and BD ' = BD ( fig . 3 ) . Join CD ' , AD ' , ED ' . By construction , AABD ' = ABD , ABC , by hypothesis . By applying the proposition ...
Page 4
... opposite sides respectively ; shew that four times the sum of the squares of BE and CF is equal to five times the square of BC . 4BE2 = 4 ( AB + AE2 ) by ( 4 ) ( fig . 4 ) , 4CF2 = 4 ( AC2 + AF " ) ; ... 4 ( BE2 + CF2 ) = 4 ( AB2 + AC2 ) ...
... opposite sides respectively ; shew that four times the sum of the squares of BE and CF is equal to five times the square of BC . 4BE2 = 4 ( AB + AE2 ) by ( 4 ) ( fig . 4 ) , 4CF2 = 4 ( AC2 + AF " ) ; ... 4 ( BE2 + CF2 ) = 4 ( AB2 + AC2 ) ...
Page 6
... opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles . ( III . 22. ) ( B ) . If a polygon of an even number of sides be inscribed in a circle , the sum of the alternate angles ...
... opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles . ( III . 22. ) ( B ) . If a polygon of an even number of sides be inscribed in a circle , the sum of the alternate angles ...
Page 10
... opposite to the equal angles are homologous . ( VI . 4. ) ( B ) . Apply this proposition to prove that the rectangle contained by the segments of any chord passing through a given point within a circle is constant . Let AB , CD ( fig ...
... opposite to the equal angles are homologous . ( VI . 4. ) ( B ) . Apply this proposition to prove that the rectangle contained by the segments of any chord passing through a given point within a circle is constant . Let AB , CD ( fig ...
Other editions - View all
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2015 |
Common terms and phrases
AC² AN.NM Arithmetic arithmetical progression axis bisects body C₁ Cambridge centre of gravity chord CHURCHILL BABINGTON circle cloth cone Conic Sections conjugate hyperbola constant curvature curve cycloid describe diameter direction directrix distance drawn Edition ellipse equations equilibrium Fellow of St fluid focus geometrical given point Hence horizontal hyperbola inches inclined inscribed John's College joining latus-rectum least common multiple Lemma length locus meet mirror move number of seconds oscillation parabola parallel parallelogram particle perpendicular plane polygon pressure prop proportional proposition prove pullies quadrilateral quantity radius ratio rays rectangle refraction right angles sewed shew sides specific gravity spherical square straight line string surface tan² tangent triangle ABC Trinity College tube V₁ vary vertex vertical W₁ weight
Popular passages
Page 4 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Page 6 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Page 11 - AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary ; if from any point C of AB, a perpendicular be drawn to AB meeting AP and .BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF.
Page 9 - IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Page 4 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.