A Supplement to the Elements of Euclid |
From inside the book
Results 1-5 of 73
Page 5
... semi - diameter equal to any given finite straight line , that exceeds the half of the straight line joining the two given points . For , let A , B be the two given points ; andjoin A , B ; and let CD be drawn bisecting AB at right ...
... semi - diameter equal to any given finite straight line , that exceeds the half of the straight line joining the two given points . For , let A , B be the two given points ; andjoin A , B ; and let CD be drawn bisecting AB at right ...
Page 93
... diameter of the less , their squares shall be , together , the double of the squares of the two semi - diameters of ... diameter AKB , of the less circle : Then PA + PB2 = 2 KA2 + 2 KP2 , KA being a semi - diameter of the less , and KP a ...
... diameter of the less , their squares shall be , together , the double of the squares of the two semi - diameters of ... diameter AKB , of the less circle : Then PA + PB2 = 2 KA2 + 2 KP2 , KA being a semi - diameter of the less , and KP a ...
Page 104
... semi - diameter KE , at its extremity E , therefore ( E. xvi . 3. ) FE touches the circle ABC . Q. E. F. PROP . X. 12. PROBLEM . To describe a circle which shall have a given semi - diameter and its centre in a given straight line , and ...
... semi - diameter KE , at its extremity E , therefore ( E. xvi . 3. ) FE touches the circle ABC . Q. E. F. PROP . X. 12. PROBLEM . To describe a circle which shall have a given semi - diameter and its centre in a given straight line , and ...
Page 105
... semi - diameter CD has been shewn to be equal to AB , which was made equal to the given straight line L. Q. E. F. PROP . XI . 13. PROBLEM . To describe a circle , the circum- ference of which shall pass through a given point , and touch ...
... semi - diameter CD has been shewn to be equal to AB , which was made equal to the given straight line L. Q. E. F. PROP . XI . 13. PROBLEM . To describe a circle , the circum- ference of which shall pass through a given point , and touch ...
Page 108
... diameter of the given circle that passes through the given point , then the solution of this latter problem is , evidently , reduced to that of the former . PROP . XIV . 17. PROBLEM . To describe two circles , each having a given semi ...
... diameter of the given circle that passes through the given point , then the solution of this latter problem is , evidently , reduced to that of the former . PROP . XIV . 17. PROBLEM . To describe two circles , each having a given semi ...
Other editions - View all
Common terms and phrases
ABē ABCD ACē aggregate angle equal arch bisect centre chord circle ABC circumference constr describe a circle describe the circle diameter distance divided draw E equi equiangular equilateral finite straight line fourth proportional given circle given finite straight given point given ratio given square given straight line greater ratio hypotenuse inscribed isosceles triangle join K less Let AB Let ABC lines be drawn manifest meet the circumference parallel to BC parallelogram polygon PROBLEM produced PROP rectangle contained rectilineal figure remaining sides required to draw rhombus right angles segment semi-diameter shewn straight line joining tangent THEOREM touch the circle trapezium wherefore xlvii xvii xxix xxvi xxviii xxxi xxxii xxxiv
Popular passages
Page 277 - 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.
Page 560 - 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'.
Page 564 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Page 178 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square...
Page 557 - ... line and the extremities of the base have the same ratio which the other sides of the triangle have to one another: and if the segments of the base...
Page 539 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Page 325 - IF an angle of a triangle be bisected by a straight line, which likewise cuts the base ; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the...
Page 550 - AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part.
Page 555 - And if the first have a greater ratio to the second, than the third has to the fourth, but the third the same ratio to the fourth, which the fifth has to the sixth...
Page 17 - ... angles equal; and conversely if two angles of a triangle are equal, two of the sides are equal. 3. If two triangles have the three sides of one equal to the three sides of the other, each to each, do you think the two triangles are alike in every respect ? 4. If two triangles have the three angles of one equal to the three angles of the other, each to each, do you think the two triangles are necessarily alike in every respect ? 5. Draw two triangles, the angles of one being equal to the angles...