## Elements of Geometry and Trigonometry |

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Page 15

From the point C

From the point C

**draw**the line CF , making with AC the right angle ACF . Now , since ACD is a straight line , the angle FCD will be a right angle ( Prop . I. Cor . 1. ) ; and since ACE is a straight line , the angle FCE will likewise be ... Page 18

... each to each , and the included angles unequal , the third sides will be unequal ; and the greater side will belong to the triangle which has the greater included angle . Make the angle CAGB4 = D ; take AG = DE , and

... each to each , and the included angles unequal , the third sides will be unequal ; and the greater side will belong to the triangle which has the greater included angle . Make the angle CAGB4 = D ; take AG = DE , and

**draw**CG . Page 21

Then , take BD equal to AC , and

Then , take BD equal to AC , and

**draw**CD . Now , in the two triangles BDC , BAC , we have BD - AC , by construction ; the angle B equal to the angle ACB , by hypothesis ; B and the side BC common : therefore , the two triangles ... Page 22

A Let us suppose that we can

A Let us suppose that we can

**draw**two perpendiculars , AB , AC . Produce either of them , as AB , till BF is equal to AB , and Ddraw FC . Then , the two triangles CAB , CBF , will be equal : for , the angles CBA , and . Page 23

A Produce the perpendicular AB till BF is equal to AB , and

A Produce the perpendicular AB till BF is equal to AB , and

**draw**FC , FD . D F First . The triangle BCF , is equal to the triangle BCA , for they have the right angle CBF CBA , the side CB common , and the side BF - BA ; hence the third ...### 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 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