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 sider |
Inni boken
Resultat 1-5 av 100
Side 9
... shewn to coincide with the point E ; wherefore the base BC shall coincide with the base EF ; because the point B coinciding with E , and C with F , if the base BC do not coincide with the base EF , the two straight lines BC and EF would ...
... shewn to coincide with the point E ; wherefore the base BC shall coincide with the base EF ; because the point B coinciding with E , and C with F , if the base BC do not coincide with the base EF , the two straight lines BC and EF would ...
Side 17
... shewn to be equal to two right angles ; wherefore the angles CEA , AED are equal to the angles AED , DEB : take away from each the common angle AED , and the remaining angle CEA is equal to the remaining angle DEB . ( ax . 3. ) In the ...
... shewn to be equal to two right angles ; wherefore the angles CEA , AED are equal to the angles AED , DEB : take away from each the common angle AED , and the remaining angle CEA is equal to the remaining angle DEB . ( ax . 3. ) In the ...
Side 21
... shewn that BA , AC are greater than BÈ , EC ; much more then are BA , AC greater than BD , DC . Again , because the exterior angle of a triangle is greater than the interior and opposite angle ; ( 1. 16. ) therefore the exterior angle ...
... shewn that BA , AC are greater than BÈ , EC ; much more then are BA , AC greater than BD , DC . Again , because the exterior angle of a triangle is greater than the interior and opposite angle ; ( 1. 16. ) therefore the exterior angle ...
Side 23
... shewn , that the angle BAC'is not equal to the angle EDF ; therefore the angle BAC is greater than the angle EDF . Wherefore , if two triangles , & c . Q.E.D. PROPOSITION XXVI . THEOREM . If two triangles have two BOOK I. PROP . XXIV ...
... shewn , that the angle BAC'is not equal to the angle EDF ; therefore the angle BAC is greater than the angle EDF . Wherefore , if two triangles , & c . Q.E.D. PROPOSITION XXVI . THEOREM . If two triangles have two BOOK I. PROP . XXIV ...
Side 29
... shewn to be equal to the angle BAC ; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB , ABC . ( ax . 2. ) Again , because the angle ACD is equal to the two angles ABC , BAC , to each of these ...
... shewn to be equal to the angle BAC ; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB , ABC . ( ax . 2. ) Again , because the angle ACD is equal to the two angles ABC , BAC , to each of these ...
Vanlige uttrykk og setninger
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
Populære avsnitt
Side 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...
Side 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.
Side 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.
Side 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.
Side 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.
Side 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.
Side 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.
Side 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.
Side 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...
Side 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...