## Elements of Geometry and Trigonometry |

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Page 17

**Hence**, the point D , falling at the same time in the two straight lines BA and CA , must fall at their intersection A :**hence**, the two triangles EDF , BAC , coincide with each other , and are therefore equal ( Ax . 13. ) . Cor . Page 23

The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the - D side BF = BA ;

The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the - D side BF = BA ;

**hence**the third sides , CF and CA are equal ( Prop . V. Cor . ) . But ABF , being a straight line ... Page 28

**Hence**AB and CD are . perpendicular to the same straight line ;**hence**they are parallel ( Prop . XVIII . ) . A R Q P B PROPOSITION XXIII . THEOREM . Two parallels are every where equally distant . Two parallels AB , CD , being CH given ... Page 29

and since DG is parallel to AB , the angle DGC is equal to BAC ;

and since DG is parallel to AB , the angle DGC is equal to BAC ;

**hence**, the angle DEF is equal to BAC ( Ax . 1. ) . Scholium . The restriction of this proposition to the case where the side EF lies in the same direction with AC ... Page 31

right angles :

right angles :

**hence**, if all the angles of a quadrilateral are equal , each of them will be a right angle ; a conclusion which sanctions the seventeenth Definition , whère the four angles of a quadrilateral are asserted to be right ...### What people are saying - Write a review

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### Common terms and phrases

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