The Elements of Euclid, containing the first six books, with a selection of geometrical problems. To which is added the parts of the eleventh and twelfth books which are usually read at the universities. By J. Martin1874 |
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Αποτελέσματα 11 - 15 από τα 43.
Σελίδα 46
... shown to be equal to BDC ; therefore , the opposite sides and angles of a parallelogram are equal to one another . Also , the diameter BC bisects it . For since AB is equal to CD and BC common , the two sides AB , BC are equal to the ...
... shown to be equal to BDC ; therefore , the opposite sides and angles of a parallelogram are equal to one another . Also , the diameter BC bisects it . For since AB is equal to CD and BC common , the two sides AB , BC are equal to the ...
Σελίδα 89
... shown , that no other point but F is the centre ; that is , 5. F is the centre of the circle ABC . Which was to be found . Cor . From this it is manifest , that if in a circle a straight line bisects another at right angles , the centre ...
... shown , that no other point but F is the centre ; that is , 5. F is the centre of the circle ABC . Which was to be found . Cor . From this it is manifest , that if in a circle a straight line bisects another at right angles , the centre ...
Σελίδα 92
... shown to be a right angle ; therefore 3. The angle FEA is equal to the angle FEB ( Ax . 1 ) , the less equal to the greater , which is impossible ; therefore 4. AC , BD do not bisect one another . Wherefore , if in a circle , & c ...
... shown to be a right angle ; therefore 3. The angle FEA is equal to the angle FEB ( Ax . 1 ) , the less equal to the greater , which is impossible ; therefore 4. AC , BD do not bisect one another . Wherefore , if in a circle , & c ...
Σελίδα 93
... shown to be equal to EC ; therefore 3. EF is equal to EG ( Ax . 1 ) , the less line equal to the greater , which is impossible . Therefore 4. E is not the centre of the circles ABC , CDG . Wherefore , if two circles , & c . Q.E.D. ...
... shown to be equal to EC ; therefore 3. EF is equal to EG ( Ax . 1 ) , the less line equal to the greater , which is impossible . Therefore 4. E is not the centre of the circles ABC , CDG . Wherefore , if two circles , & c . Q.E.D. ...
Σελίδα 94
... shown to be equal to FC ; therefore 3. FE is equal to FB ( Ax . 1 ) , the less line equal to the greater , which is impossible ; therefore 4. F is not the centre of the circles ABC , CDE . Therefore , if two circles , & c . Q.E.D. ...
... shown to be equal to FC ; therefore 3. FE is equal to FB ( Ax . 1 ) , the less line equal to the greater , which is impossible ; therefore 4. F is not the centre of the circles ABC , CDE . Therefore , if two circles , & c . Q.E.D. ...
Άλλες εκδόσεις - Προβολή όλων
The Elements of Euclid, Containing the First Six Books, with a Selection of ... Euclides Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2016 |
Συχνά εμφανιζόμενοι όροι και φράσεις
AC is equal adjacent angles angle ABC angle ACB angle BAC angle BCD angle DEF angle EDF angle equal base BC bisected centre circle ABC circumference constr Demonstration diameter double equal angles equal to F equiangular equilateral triangle equimultiples ex æquali exterior angle fourth given circle given point given straight line gnomon greater ratio inscribed less Let ABC Let the straight linear units meet multiple opposite angle parallel to BC parallelogram perpendicular plane polygon produced proportionals Q.E.D. PROPOSITION quadrilateral rectangle contained remaining angle right angles segment semicircle similar square on AC straight line AB straight line BC straight line drawn three straight lines tiple touches the circle triangle ABC triangle DEF twice the rectangle wherefore
Δημοφιλή αποσπάσματα
Σελίδα 1 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.
Σελίδα 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...
Σελίδα 232 - If two triangles, which have two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to one another, the remaining sides shall be in a straight line. Let...
Σελίδα 112 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.
Σελίδα 209 - ... triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.
Σελίδα 269 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.
Σελίδα 199 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Σελίδα 23 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Σελίδα 63 - 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...
Σελίδα 32 - ... 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.