A Companion to Wrigley's Collection of Examples: Being Illustrations of Mathematical Processes and Methods of Solution |
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Page 96
... perpendicular to BC . 4. Let BAC be the triangle , AD the line bisecting the angle BAC , and the base BC . Produce AD to E , and make DE AD , and join BE . = Now , in the triangles ADC , BDE , we have AD = DE ( by construction ) ; BDBC ...
... perpendicular to BC . 4. Let BAC be the triangle , AD the line bisecting the angle BAC , and the base BC . Produce AD to E , and make DE AD , and join BE . = Now , in the triangles ADC , BDE , we have AD = DE ( by construction ) ; BDBC ...
Page 97
... perpendicular to CD , and produce it to E , making FE equal to AF , and join BE cutting CD in G. Join also AG . Then AG and BG make equal angles with CD . For since AF is equal to FE , and FG is common to the two triangles AGF , EGF ...
... perpendicular to CD , and produce it to E , making FE equal to AF , and join BE cutting CD in G. Join also AG . Then AG and BG make equal angles with CD . For since AF is equal to FE , and FG is common to the two triangles AGF , EGF ...
Page 98
... perpendicular at D passes through G. B 17. From B the right angle of the triangle ABC let BE be drawn perpendicular to the hypo- thenuse , and BD bisecting the angle ABC : the angle EBD is half the difference of the angles BAC , BCA ...
... perpendicular at D passes through G. B 17. From B the right angle of the triangle ABC let BE be drawn perpendicular to the hypo- thenuse , and BD bisecting the angle ABC : the angle EBD is half the difference of the angles BAC , BCA ...
Page 99
... perpendicular to BC . Then ( Eucl . II . 12 ) ABAD + DB2 + 2BD × DE , and ( Eucl . II . 13 ) с B E ACAD2 + DC2 - 2 CD x DE = AD3 + DB2 — 2BD × DE , - whence AB + AC2 = 2AD + 2BD3 . 32. Let ABC be any triangle ; and let BP , CQ be the ...
... perpendicular to BC . Then ( Eucl . II . 12 ) ABAD + DB2 + 2BD × DE , and ( Eucl . II . 13 ) с B E ACAD2 + DC2 - 2 CD x DE = AD3 + DB2 — 2BD × DE , - whence AB + AC2 = 2AD + 2BD3 . 32. Let ABC be any triangle ; and let BP , CQ be the ...
Page 101
... perpendicular to AB , CD , and therefore bisecting them : then difference of AE , EB = 2EF , and difference of CE , ED = 2EG : but AB2 ~ CD2 = 4BF2 — 4CG2 = 40G ~ 40F = 4EF3 ~ 4EG = ( 2EF ) 2 ~ ( 2EG ) ' = = - = ( AE — EB ) 2 ~ ( CE ...
... perpendicular to AB , CD , and therefore bisecting them : then difference of AE , EB = 2EF , and difference of CE , ED = 2EG : but AB2 ~ CD2 = 4BF2 — 4CG2 = 40G ~ 40F = 4EF3 ~ 4EG = ( 2EF ) 2 ~ ( 2EG ) ' = = - = ( AE — EB ) 2 ~ ( CE ...
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A Companion to Wrigley's Collection of Examples: Being Illustrations of ... John Thompson Platts,Alfred Wrigley No preview available - 2016 |
Common terms and phrases
a+b+c a²+b² a²b³ AB² ABCD AC² angle ABC axis BC² beam centre of gravity chord circle co-ordinates coefficient cos² cos³ cosec² curve diameter draw ellipse equal equilateral Eucl Hence hyperbola Join latus rectum Let ABC middle point MULTINOMIAL THEOREMS parabola parallel perpendicular plane point of contact point of intersection r₁ radius required locus right angle roots S₁ sec² segment sides Similarly sin² sin³ square straight line tan² tangent term transformed equation triangle ABC values vertical weight whence X₁