## Elements of Geometry |

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Page 2

The side

The side

**opposite**to the right angle is called the hypothenuse . Fig . 10. Thus ABC ( fig . 10 ) is a triangle right - angled at A , and the side BC is the hypothenuse . 17. Among quadrilateral figures we distinguish , Fig . 11 . Page 6

It may be demonstrated , in like manner , that the angle ACE is equal to its

It may be demonstrated , in like manner , that the angle ACE is equal to its

**opposite**angle BCD . 35. Scholium . The four angles , formed about a point by two straight lines which cut each other , are together equal to four right angles ... Page 9

It may be remarked , that equal angles are

It may be remarked , that equal angles are

**opposite**to equal sides ; thus , the equal angles A and D are**opposite**to the equal sides B C and EF . THEOREM . 45. In an isosceles triangle , the angles**opposite**to the equal sides are equal ... Page 10

In a triangle that is not isosceles , any one of its sides may be taken indifferently for a base ; and then its vertex is that of the

In a triangle that is not isosceles , any one of its sides may be taken indifferently for a base ; and then its vertex is that of the

**opposite**angle . In an isosceles triangle , the base is that side which is not equal to one of the ... Page 11

The two oblique lines AC , AE , which meet the line DE on

The two oblique lines AC , AE , which meet the line DE on

**opposite**sides of the perpendicular , and at equal distances BC , BE , from it , are equal to one another ; 3. Of any two oblique lines AC , AD , or AE , AD , that which is more ...### 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