## Elements of Geometry and Trigonometry |

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Page 24

Then , in the two triangles BAG , DEF , the angles B and E are equal , being right angles , the side BA - ED by hypothesis , and the side BG = EF by construction :

Then , in the two triangles BAG , DEF , the angles B and E are equal , being right angles , the side BA - ED by hypothesis , and the side BG = EF by construction :

**consequently**, AG - DF ( Prop . V. Cor . ) . But , by hypothesis AC - DF ... Page 32

angles , is equal to twice as many right angles as the polygon has sides , and

angles , is equal to twice as many right angles as the polygon has sides , and

**consequently**, equal to the sum of the interior angles plus the exterior angles . Taking from each the sum of the interior angles , and there remains the ... Page 33

moreover , the side DB is common , and the side AB = DC ; hence the triangle ABD is equal to the triangle DBC ( Prop . V. ) ; therefore , the side AD is equal to BC , the angle ADB DBC , and

moreover , the side DB is common , and the side AB = DC ; hence the triangle ABD is equal to the triangle DBC ( Prop . V. ) ; therefore , the side AD is equal to BC , the angle ADB DBC , and

**consequently**AD is parallel to BC ; hence the ... Page 40

M N Let M and N be any two magnitudes , and . and be like m m parts of each : then will Let then will and m N For , it is obvious that Mx ( N ) = Nx ( M ± ( M + M ) since m N.M

M N Let M and N be any two magnitudes , and . and be like m m parts of each : then will Let then will and m N For , it is obvious that Mx ( N ) = Nx ( M ± ( M + M ) since m N.M

**Consequently**, the four quanm each is cqual to M.N ± tities ... Page 41

In all cases , the same chord FG belongs to two arcs , FGH , FEĠ , and

In all cases , the same chord FG belongs to two arcs , FGH , FEĠ , and

**consequently**also to two segments : but the smaller one is always meant , unless the contrary is expressed . D * An inscribed triangle is one which , like BAC ...### What people are saying - Write a review

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ABCD adjacent altitude base become Book called centre chord circle circumference circumscribed common cone consequently contained Cosine Cotang cylinder described determine diameter difference distance divided draw drawn equal equations equivalent expressed extremities faces feet figure follows formed four frustum give given gles greater half hence homologous hypothenuse included inscribed intersection less let fall logarithm manner means measured meet middle multiplied number of sides opposite parallel parallelogram pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quadrant quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment sides similar Sine solid solid angle sphere spherical triangle square straight line suppose taken Tang tangent THEOREM third triangle triangle ABC unit vertex whole