Elements of Geometry: On the Basis of Dr. Brewster's Legendre : to which is Added a Book on Proportion, with Notes and Illustrations |
From inside the book
Results 1-5 of 25
Page 19
Magnitudes , whether lines , surfaces , or solids , which coincide throughout their whole extent , i . e . which exactly fill the same space , are equal . PROPOSITION I. THEOREM . Every straight line CD , which BOOK I. 19 Explanation of ...
Magnitudes , whether lines , surfaces , or solids , which coincide throughout their whole extent , i . e . which exactly fill the same space , are equal . PROPOSITION I. THEOREM . Every straight line CD , which BOOK I. 19 Explanation of ...
Page 21
Two straight lines , which have two points A and B common to both , coincide with each other throughout their whole extent , and form one and the same straight line . First , since the points A and B are F common to both lines , it is ...
Two straight lines , which have two points A and B common to both , coincide with each other throughout their whole extent , and form one and the same straight line . First , since the points A and B are F common to both lines , it is ...
Page 24
But by hypothesis , DF is equal to AC ; therefore the point F will fall on C , and the third side EF must ex . actly coincide with the third side BC ; ( Ax . 12 ; ) therefore the triangle DEF coincides with the triangle ABC throughout ...
But by hypothesis , DF is equal to AC ; therefore the point F will fall on C , and the third side EF must ex . actly coincide with the third side BC ; ( Ax . 12 ; ) therefore the triangle DEF coincides with the triangle ABC throughout ...
Page 25
Hence , the point D , occurring at the same time in the two straight lines BA and CA , must fall on their intersection A ; hence the two triangles ABC , DEF , coincide with each other , and are consequently equal . ( Ax . 13. ) ...
Hence , the point D , occurring at the same time in the two straight lines BA and CA , must fall on their intersection A ; hence the two triangles ABC , DEF , coincide with each other , and are consequently equal . ( Ax . 13. ) ...
Page 51
K For , since the diameters AB , EF are equal , the semicircle AMDB may be applied exactly to the semicircle ENGF , and the curve line AMDB will coincide entirely with the curve line ENGF . But the part AMD is equal to the part ENG by ...
K For , since the diameters AB , EF are equal , the semicircle AMDB may be applied exactly to the semicircle ENGF , and the curve line AMDB will coincide entirely with the curve line ENGF . But the part AMD is equal to the part ENG by ...
What people are saying - Write a review
We haven't found any reviews in the usual places.
Common terms and phrases
ABCD Abridgment Academies adjacent altitude antecedent base called centre chord circle circumference circumscribed coincide common cone consequently construction contained convex surface cylinder Day's Algebra demonstration described diagonal diameter difference distant divided draw drawn equal equivalent expressed extremities faces fall figure four frustum geometry given greater half hence hypothesis inscribed interior intersection join lateral less magnitudes manner mean measured meet multiplied number of sides opposite parallel parallelogram parallelopipedon pass perimeter perpendicular plane plane MN polygon principles prism PROBLEM Prop proportional PROPOSITION proved pyramid quantity radii radius ratio reason rectangle regular respectively right-angles Scholium schools segment sides similar solid angle sphere square straight line suppose surface taken tangent THEOREM third triangle triangle ABC vertex whole
Popular passages
Page 196 - THEOREM. Every section of a sphere, made by a plane, is a circle.
Page 176 - AT into equal parts .Ax, xy, yz, &c., each less than Aa, and let k be one of those parts : through the points of division pass planes parallel to the plane of the bases : the corresponding sections formed by these planes in the two pyramids will be respectively equivalent, namely, DEF to def, GHI to ghi, &c.
Page 125 - AB as a diameter, describe a semicircle : at the extremity of the diameter draw the tangent AD, equal to the side of the square C ; through the point D and the centre O draw the secant DF ; then will DE and DF be the adjacent sides of the rectangle required. For...
Page 229 - The area of the circle, we infer therefore, is equal to 3.1415926. Some doubt may exist perhaps about the last decimal figure, owing to errors proceeding from the parts omitted ; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place. Since the...
Page 118 - B, may be found in the same manner, for it will be the same as a fourth proportional to the three lines A, B, B. PROBLEM IIL To find a mean proportional between two given lines A and B.
Page 176 - DEF, def, are equivalent; for like reasons, the third exterior prism GHI-K and the second interior prism ghi-d are equivalent; the fourth exterior and the third interior ; and so on, to the last in each series. Hence all the exterior prisms of the pyramid...
Page 46 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Page 220 - Let it be granted that a straight line may be drawn from any one point to any other point.
Page 101 - In every triangle, the square of the side subtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle.
Page 227 - The surface of a regular inscribed polygon, and that of a similar polygon circumscribed, being given ; to find the surfaces of the regular inscribed and circumscribed polygons having double the number of sides. Let AB be a side of the given inscribed polygon ; EF, parallel to AB, a side of the circumscribed polygon ; C the centre of the circle. If the chord AM and the tangents AP, BQ, be drawn, AM...