A Supplement to the Elements of EuclidJ. Smith, 1825 - 582 sider |
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Side 35
... XXXIV . 48. PROBLEM . To describe a triangle which shall have its three sides , taken together , equal to a given finite straight line , and its three angles equal to three given angles , each to each ; the three given angles being ...
... XXXIV . 48. PROBLEM . To describe a triangle which shall have its three sides , taken together , equal to a given finite straight line , and its three angles equal to three given angles , each to each ; the three given angles being ...
Side 44
... XXXIV . 1. ) FE = BD = BC ( constr . ) ; also , since ( constr . ) the FBD is a right angle , the < BFE is , also , ( E. xxix . 1. ) a right angle . PROP . XLII . 57. THEOREM . The diameters of a parallelogram bisect each other . Let AB ...
... XXXIV . 1. ) FE = BD = BC ( constr . ) ; also , since ( constr . ) the FBD is a right angle , the < BFE is , also , ( E. xxix . 1. ) a right angle . PROP . XLII . 57. THEOREM . The diameters of a parallelogram bisect each other . Let AB ...
Side 45
... XXXIV . 1. ) and ( E. xxix . 1. ) the LEAD of the AAED , LEBC , of the ABEC , and the LEDA LECB ; therefore ( E ... XXXIV . 1. ) the A = < C ; therefore ( E.1v . 1. ) FE = GH : Again since ( E. XXXIV.1 . ) AB = DC , and AD = BC , and ...
... XXXIV . 1. ) and ( E. xxix . 1. ) the LEAD of the AAED , LEBC , of the ABEC , and the LEDA LECB ; therefore ( E ... XXXIV . 1. ) the A = < C ; therefore ( E.1v . 1. ) FE = GH : Again since ( E. XXXIV.1 . ) AB = DC , and AD = BC , and ...
Side 46
... XXXIV . 1. ) the EDH = LFBG ; therefore ( E. IV . 1. ) EH = FG ; and it has been proved that EF = HG ; therefore ( Supp . xviii . 1. ) EFGH is a parallelogram . PROP . XLIV . 59. THEOREM . If any number of parallelograms be inscribed in ...
... XXXIV . 1. ) the EDH = LFBG ; therefore ( E. IV . 1. ) EH = FG ; and it has been proved that EF = HG ; therefore ( Supp . xviii . 1. ) EFGH is a parallelogram . PROP . XLIV . 59. THEOREM . If any number of parallelograms be inscribed in ...
Side 48
... XXXIV . 1. ) AD = BC ; and AB is common to the two AS ABC , BAD , and the LABC = LBAD ; therefore ( E. iv . 1. ) AC = BD . PROP . XLVII . 63. PROBLEM . To inscribe a square in a given equilateral four - sided figure . Let ABCD be the ...
... XXXIV . 1. ) AD = BC ; and AB is common to the two AS ABC , BAD , and the LABC = LBAD ; therefore ( E. iv . 1. ) AC = BD . PROP . XLVII . 63. PROBLEM . To inscribe a square in a given equilateral four - sided figure . Let ABCD be the ...
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Vanlige uttrykk og setninger
AB² ABCD AC² 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
Populære avsnitt
Side 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.
Side 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'.
Side 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.
Side 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...
Side 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...
Side 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.
Side 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...
Side 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.
Side 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...
Side 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...