Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 688.
Wednesday, November 16, 2011
Problem 688: Triangle, Angles, 10, 20, 30, 40, 60 Degrees, Measure, Mind Map, Polya
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http://img403.imageshack.us/img403/6706/sfvdf.png
ReplyDeleteExtend AB from B till F so that < EFD = 30°. Since < CEF = 50°, then < CDF = 50° and DF ⊥ BC, whereat < BDF = 30°, and finally from the cyclic quadrilateral CDEF we have x + 30° = 50° and x = 20°.
Extend AB till F so that AF=AC. This way DCFE is an isosceles trapezoid(CD=CF=EF and <DCF=<CFE=80). <AED =80=<EBD+<BDE. So <BDE=20.
ReplyDeleteDid you both used AD=AE which isn't given or how can someone conclude EF=CD?
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