## Elements of Geometry and Trigonometry |

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Page 24

Then , in the two triangles BAG , DEF , the angles R 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 R 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 40

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

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

**Consequently**, the four quan- m ... Page 41

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

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

**consequently**also to two segments : but the smaller one is always meant unless the contrary is expressed . An inscribed triangle is one which , like BAC , D * 6 BOOK ... Page 44

... the triangles ACD , EOG , will have all their sides equal , each to each , namely , AC = EO , CD = OG , and AD = EG ; hence the triangles are themselves equal ; and ,

... the triangles ACD , EOG , will have all their sides equal , each to each , namely , AC = EO , CD = OG , and AD = EG ; hence the triangles are themselves equal ; and ,

**consequently**, the angle ACD is equal EOG ( Book I. Prop . X. ) .### 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 construction contained corresponding cosine Cotang cylinder described diameter difference distance divided draw drawn equal equation equivalent evident expressed extremities fall figure follows formed formulas four frustum give given gles greater half hence homologous included inscribed intersection less likewise logarithm manner means measured meet middle multiplied opposite parallel parallelogram parallelopipedon pass perpendicular plane polygon prism PROBLEM Prop proportional PROPOSITION pyramid quantities radii radius ratio reason rectangle regular remaining right angles Scholium segment shown sides similar sine solid solid angle sphere spherical triangle square straight line Suppose surface taken tang tangent THEOREM third triangle triangle ABC vertex whole