## Elements of Geometry |

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Page 24

46 ) , & c . , drawn from the centre to the circumference , is called a

46 ) , & c . , drawn from the centre to the circumference , is called a

**radius**or semidiameter , and every straight line , as AB , which passes through the centre , and is terminated each way by the circumference , is called a diameter ... Page 26

The

The

**radius**AC ( fig . 50 ) being equal to the**radius**EO , and the arc AMD equal to the arc ENG , the chord AD will be equal to the chord EG . For , the diameter AB being equal to the diameter EF , the semicircle AMDB may be applied ... Page 27

The

The

**radius**CG ( fig . 51 ) , perpendicular to a chord AB , Fig . 61 . bisects this chord and the arc subtended by it AGB . Demonstration . Draw the radii CA , CB ; these radii are , with respect to the perpendicular CD , two equal ... Page 28

... therefore the circumference , described from the centre 0 with the

... therefore the circumference , described from the centre 0 with the

**radius**OB , will pass through the three points A , B , C. , It is thus proved , that the circumference of a circle may be made to pass through any three given points ... Page 29

54 .

54 .

**radius**AC , is a tangent to the circumference . Demonstration . Since every oblique line CE is greater than the perpendicular CA ( 52 ) , the point E is without the circle , and the line BD has only the point A in common with the ...### What people are saying - Write a review

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ABC fig ABCD adjacent altitude applied base called centre chord circ circle circumference circumscribed common cone consequently construction contained convex surface Corollary cylinder Demonstration described diameter difference distance divided draw drawn entire equal equivalent example extremities faces feet figure follows formed four give given greater half hence inclination inscribed intersection isosceles join less let fall manner mean measure meet moreover multiplied namely opposite parallel parallelogram parallelopiped pass perimeter perpendicular plane plane angles polyedron polygon prism PROBLEM produced proportional proposition pyramid radii radius ratio reason rectangle regular polygon respect right angles Scholium sector segment similar solid angle Solution sphere spherical square straight line suppose surface taken tangent THEOREM third triangle ABC vertex vertices whence