Key to Exercises in EuclidMacmillan and Company, 1880 - 114 sider |
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Side 1
... circle ; with Band radius equal to DE , describe another circle ; let these circles at C. Join AC and BC ; then ABC will be the triangle required . The given point and the vertex of the constructed triangle both fall on the ...
... circle ; with Band radius equal to DE , describe another circle ; let these circles at C. Join AC and BC ; then ABC will be the triangle required . The given point and the vertex of the constructed triangle both fall on the ...
Side 11
... circle at K. Join CK , HK , AH , CH . The angle CAB is one third of two right angles ; therefore the angle CAK is two thirds of two right angles . Also , each of the angles CAB , HAB being one third of two right angles , the whole angle ...
... circle at K. Join CK , HK , AH , CH . The angle CAB is one third of two right angles ; therefore the angle CAK is two thirds of two right angles . Also , each of the angles CAB , HAB being one third of two right angles , the whole angle ...
Side 12
... circle with DE , and thus two solutions apparently : but it will be found on examination that there is only one distinct solution . 66. Take a straight line AD equal to the given difference of the sides . At the point D draw a straight ...
... circle with DE , and thus two solutions apparently : but it will be found on examination that there is only one distinct solution . 66. Take a straight line AD equal to the given difference of the sides . At the point D draw a straight ...
Side 15
... circle meeting CD at F ; join EF . Through O draw a straight line parallel to EF , meeting AB at G , and CD at H. Then EGHF is a parallelogram , by construction ; therefore GH is equal to EF , by I. 34 : thus GH is equal to the given ...
... circle meeting CD at F ; join EF . Through O draw a straight line parallel to EF , meeting AB at G , and CD at H. Then EGHF is a parallelogram , by construction ; therefore GH is equal to EF , by I. 34 : thus GH is equal to the given ...
Side 25
... circle cutting BE at F. perpendicular to AB : then AB shall be divided at D in the manner required . For it may be shewn as in II . 9 that FD is equal to DB . Also the squares on AD and DF are equal to the square on so that the squares ...
... circle cutting BE at F. perpendicular to AB : then AB shall be divided at D in the manner required . For it may be shewn as in II . 9 that FD is equal to DB . Also the squares on AD and DF are equal to the square on so that the squares ...
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ABCD AC and BD AC is equal angle ABC angle ACB angle BAC angle CAB bisects the angle Cambridge centre chord circle described circumference Crown 8vo describe a circle diameter draw a straight Edition equal to BC equal to four equal to half equal to twice equiangular Euclid Exercise 176 Extra fcap fcap fore given circle given point given straight line greater half a right half the angle Hence Let ABC meeting BC middle point Owens College parallel to BC parallelogram perpendicular plane Professor quadrilateral radius equal rectangle rhombus right angles segment shewn side BC square on BC straight line parallel suppose tangent touch the circle triangle ABC twice the angle twice the rectangle twice the squares
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