## 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

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**hence**the A sum of the two angles ACD , DCB , must be equal to two right - angles . E D -B In the same manner it may be proved that every straight line which meets another , makes with it two adjacent angles , whose sum is equal to two ... Page 25

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**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 , coin- cide with each other , and are consequently equal . ( Ax . 13. )**Hence**... Page 29

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**hence**the angle B = C ; and since the side AB = BC , there- fore the angle A = C ;**hence**the angle A = B ; ( Ax . 1 ; ) therefore the three an- gles A , B , C , are equal to each other . B A C Scholium . The equality of the triangles ... Page 30

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**hence**the side AB is not greater than AC . In like manner it may be proved that AB is not less than AC ,**hence**there is no inequality between the sides AB , AC ; therefore the triangle ABC is isosceles . ( Def . 17. )**Hence**, If two ... Page 31

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**hence**the angle BCF = BCA . ( Prop . 5. ) But BCA is a right - angle by hypothesis ; there- fore BCF must also be a right - angle . Now since the adjacent angles BCA , BCF , are together equal to two right - angles , the line ACF must ...### Common terms and phrases

Abridgment of Day's adjacent angles allel altitude angle ACB angle BAC antecedent base ABCD bisect centre chord circ circle circumference circumscribed polygon common cone consequently convex surface couplets cylinder Day's Algebra diagonal diameter divided draw drawn equal and parallel equal angles equally distant equiangular equilateral triangles equivalent four magnitudes frustum geometry greater half homologous sides hypothenuse hypothesis inscribed polygon interior angles intersection let fall manner mean proportional measured number of sides oblique lines opposite parallelogram pendicular perimeter perpendicular plane angles plane MN polyedron polygon ABCDE President Day prism PROBLEM Prop PROPOSITION XI pyramid SABCDE quadrilateral quantity radii radius ratio rectangle regular polygon respectively equal SABC Scholium Schools and Academies segment similar solid angle sphere square described straight line tangent THEOREM Thomson trapezium triangle ABC triangular prism vertex Yale College

### 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...