A Companion to Wrigley's Collection of Examples: Being Illustrations of Mathematical Processes and Methods of Solution |
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Page 96
... Produce AD to E , and make DEAD , and join BE . Now , in the triangles ADC , BDE , we have AD = DE ( by construction ) ; BDBC ( by hypothesis ) ; and = LBDE LADC ( Euc . 1. 15 ) . Therefore ( Euc . 1. 4 ) BE = AC , and angle BED = angle ...
... Produce AD to E , and make DEAD , and join BE . Now , in the triangles ADC , BDE , we have AD = DE ( by construction ) ; BDBC ( by hypothesis ) ; and = LBDE LADC ( Euc . 1. 15 ) . Therefore ( Euc . 1. 4 ) BE = AC , and angle BED = angle ...
Page 97
... 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 , and the included angles AFG ...
... 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 , and the included angles AFG ...
Page 106
... produce it to D. Join OD ; and through A draw OAE ; and draw BE parallel to OD , cutting OAE in E. E is the cen- tre of the circle required . Since ( Eucl . 1. 29 ) the angle ODA is equal to ABE , and OAD to BAE , therefore the ...
... produce it to D. Join OD ; and through A draw OAE ; and draw BE parallel to OD , cutting OAE in E. E is the cen- tre of the circle required . Since ( Eucl . 1. 29 ) the angle ODA is equal to ABE , and OAD to BAE , therefore the ...
Page 107
... produce BA to D , mak- ing AD = AB . From D draw DE parallel to AC , meeting the circle in E ; join BE , cut- ting AC in F ; BF will be a mean proportional between AF D and FC . For ( Eucl . VI . 2 ) BF : FE :: BA : AD , E. and since ...
... produce BA to D , mak- ing AD = AB . From D draw DE parallel to AC , meeting the circle in E ; join BE , cut- ting AC in F ; BF will be a mean proportional between AF D and FC . For ( Eucl . VI . 2 ) BF : FE :: BA : AD , E. and since ...
Page 109
... produced in S and T. Draw PM at right angles to AC and PN to BC . Then triangle SCT : triangle ACB = CS . CT : CA. CB , S A M also CS : CP = CP : CM , and CT : CP = CP : CN ; T C N B therefore CS.ST : CP CP : CN . CM CP : CM . MP ; = CM ...
... produced in S and T. Draw PM at right angles to AC and PN to BC . Then triangle SCT : triangle ACB = CS . CT : CA. CB , S A M also CS : CP = CP : CM , and CT : CP = CP : CN ; T C N B therefore CS.ST : CP CP : CN . CM CP : CM . MP ; = CM ...
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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