## Elements of Geometry and Trigonometry |

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Page 13

10. All right angles are equal to each other . 11 From one point to another only one straight line can be drawn . 12. Through the same point , only one straight line can be drawn which shall be parallel to a

10. All right angles are equal to each other . 11 From one point to another only one straight line can be drawn . 12. Through the same point , only one straight line can be drawn which shall be parallel to a

**given**line . 13. Page 21

It the angle CB , then the side AB = AC ( Prop . XII . ) ; which is also contrary to the supposition . Therefore , when AB > AC , the angle C must be greater than B. PROPOSITION XIV . THEOREM . From a

It the angle CB , then the side AB = AC ( Prop . XII . ) ; which is also contrary to the supposition . Therefore , when AB > AC , the angle C must be greater than B. PROPOSITION XIV . THEOREM . From a

**given**point , BOOK I. 21. Page 22

From a

From a

**given**point , without a straight line , only one perpendicular can be drawn to that line . Let A be the point , and DE the**given**line . A Let us suppose that we can draw two perpendiculars , AB , AC . Page 23

Let A be the

Let A be the

**given**point , DE the**given**line , AB the perpendicular , and AD , AC , AE , the oblique lines . 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 ... Page 24

Let AB be the

Let AB be the

**given**straight line , C the middle point , and ECF the perpendicular . First , Since AC = CB , the two oblique lines AD , DB , are equally distant from the perpendicular , and therefore equal ( Prop . XV . ) .### 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