Key to Exercises in EuclidMacmillan and Company, 1880 - 114 sider |
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... Keys , to present the solutions in a simple natural manner , in order to meet the difficulties which are most likely to arise , and to render the work intelligible and instructive . November , 1880 . PAGE 1 3 6 8 14 18 22 24 26.
... Keys , to present the solutions in a simple natural manner , in order to meet the difficulties which are most likely to arise , and to render the work intelligible and instructive . November , 1880 . PAGE 1 3 6 8 14 18 22 24 26.
Side 2
... meet AB at E. It may be shewn as before that AE is equal to BE , and that the angles at E are right angles . 13. Let AB be the given straight line , C and D the two given points , Join CD and bisect it at E. From E draw a straight line ...
... meet AB at E. It may be shewn as before that AE is equal to BE , and that the angles at E are right angles . 13. Let AB be the given straight line , C and D the two given points , Join CD and bisect it at E. From E draw a straight line ...
Side 3
... meet at the point E , and make the angle AEB equal to the angle CED , and the angle BEC equal to the angle DEA : then shall AE and EC be in one straight line , and also BE and ED in one straight line . By I. 15 , Cor . 2 the four angles ...
... meet at the point E , and make the angle AEB equal to the angle CED , and the angle BEC equal to the angle DEA : then shall AE and EC be in one straight line , and also BE and ED in one straight line . By I. 15 , Cor . 2 the four angles ...
Side 4
... meet AB at D. Then the triangle ACD is isosceles , by I. 6. Also as the angle ACB is equal to the sum of the angles A and B , and the angle ACG is equal to the angle A , the angle BCG is equal to the angle B : therefore the triangle BCD ...
... meet AB at D. Then the triangle ACD is isosceles , by I. 6. Also as the angle ACB is equal to the sum of the angles A and B , and the angle ACG is equal to the angle A , the angle BCG is equal to the angle B : therefore the triangle BCD ...
Side 5
... meet , produced if necessary , at 0. Through O draw a straight line bisecting the angle between CD and EF ; and let AB , produced if necessary , meet this straight line at K ; then K will be such a point as is required : for the ...
... meet , produced if necessary , at 0. Through O draw a straight line bisecting the angle between CD and EF ; and let AB , produced if necessary , meet this straight line at K ; then K will be such a point as is required : for the ...
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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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