## Elements of Geometry |

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Page 12

But AF < AC + CF ( 40 ) , and AB

But AF < AC + CF ( 40 ) , and AB

**half**of AF is less than AC**half**of AC + CF ; that is , the perpendicular is less than any one of the oblique lines . 2. If BE = BC , then , as AB is common to the two triangles ABE , ABC , and the right ... Page 18

We are able , therefore , in this manner , to take successively the

We are able , therefore , in this manner , to take successively the

**half**, the fourth , the eighth , & c . , of the angle GFM , and the lines which form these divisions meet the line AB in points more and more distant , but easily ... Page 26

In the same circle , or in equal circles , if the arc be less than

In the same circle , or in equal circles , if the arc be less than

**half**a circumference , the greater arc is subtended by the greater chord ; and , conversely , the greater chord is subtended by the greater arc . Demonstration . Page 29

The right - angled triangles CAF , DCG , have the hypothenuses CA , CD , equal ; moreover , the side AF , the

The right - angled triangles CAF , DCG , have the hypothenuses CA , CD , equal ; moreover , the side AF , the

**half**of AB , is equal to the side DG , the**half**of DE ; the triangles , then , are equal ( 56 ) , and consequently the third ... Page 35

64 measure the

64 measure the

**half**of the arc BD comprehended between its sides . Demonstration . Let us suppose , in the first place , that the centre of the circle is situated in the angle BAD ( fig . 64 ) ; we Fig .### What people are saying - Write a review

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### Common terms and phrases

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