Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with Explanatory Notes; a Series of Questions on Each Book; and a Selection of Geometrical Exercises from the Senate-house and College Examination Papers, with Hints, &c. Designed for the Use of the Junior Classes in Public and Private Schools. the first six books, and the portions of the eleventh and twelfth books read at CambridgeLongman, Green, Longman, Roberts, and Green, 1868 - 410 pages |
From inside the book
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Page 11
... join DC . Then , in the triangles DBC , ABC , because DB is equal to AC , and BC is common to both triangles , the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; ( hyp ...
... join DC . Then , in the triangles DBC , ABC , because DB is equal to AC , and BC is common to both triangles , the two sides DB , BC are equal to the two sides AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; ( hyp ...
Page 13
... join DE ; on the side of DE remote from A , describe the equilateral triangle DEF ( 1. 1. ) , and join AF . Then the straight line AF shall bisect the angle BAC . Because AD is equal to AE , ( constr . ) and AF is common to the two ...
... join DE ; on the side of DE remote from A , describe the equilateral triangle DEF ( 1. 1. ) , and join AF . Then the straight line AF shall bisect the angle BAC . Because AD is equal to AE , ( constr . ) and AF is common to the two ...
Page 15
... join CH . • Then the straight line CH drawn from the given point C , shall be perpendicular to the given straight line AB . Join FC , and CG . Because FH is equal to HG , ( constr . ) and HC is common to the triangles FHC , GHC ; the ...
... join CH . • Then the straight line CH drawn from the given point C , shall be perpendicular to the given straight line AB . Join FC , and CG . Because FH is equal to HG , ( constr . ) and HC is common to the triangles FHC , GHC ; the ...
Page 17
... join BE ; produce BE to F , making EF equal to BE , ( 1. 3. ) and join FC . Because AE is equal to EC , and BE to BOOK I. PROP . XV , XVI . 17.
... join BE ; produce BE to F , making EF equal to BE , ( 1. 3. ) and join FC . Because AE is equal to EC , and BE to BOOK I. PROP . XV , XVI . 17.
Page 30
... join the extremities of two equal and parallel straight lines towards the same parts , are also themselves equal and ... Join BC . D Then because AB is parallel to CD , and BC meets them , therefore the angle ABC is equal to the ...
... join the extremities of two equal and parallel straight lines towards the same parts , are also themselves equal and ... Join BC . D Then because AB is parallel to CD , and BC meets them , therefore the angle ABC is equal to the ...
Common terms and phrases
A₁ ABCD AC is equal Algebraically angle ABC angle ACB angle BAC Apply Euc base BC chord circle ABC constr demonstrated describe a circle diagonals diameter divided double draw equal angles equiangular equilateral triangle equimultiples Euclid exterior angle Geometrical given circle given line given point given straight line gnomon greater hypotenuse inscribed intersection isosceles triangle less Let ABC line BC lines be drawn multiple opposite angles parallelogram parallelopiped pentagon perpendicular plane polygon problem produced Prop proportionals proved Q.E.D. PROPOSITION quadrilateral figure radius ratio rectangle contained rectilineal figure remaining angle right angles right-angled triangle segment semicircle shew shewn similar similar triangles solid angle square on AC tangent THEOREM touch the circle trapezium triangle ABC twice the rectangle vertex vertical angle wherefore
Popular passages
Page 6 - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...
Page 118 - Guido, with a burnt stick in his hand, demonstrating on the smooth paving-stones of the path, that the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.
Page 2 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Page 317 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Page 90 - If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC.
Page 88 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Page 30 - ... twice as many right angles as the figure has sides ; therefore all the angles of the figure together with four right angles, are equal to twice as many right angles as the figure has sides.
Page 9 - THE angles at the base of an isosceles triangle are equal to one another : and, if the equal sides be produced, the angles upon the other side of the base shall be equal.
Page 22 - IF two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other ; the base of that which has the greater angle shall be greater than the base of the other...
Page 92 - If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. Let the straight line AB be divided into two equal parts...