Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 665.
http://img33.imageshack.us/img33/974/problem665.pngDenote (XYZ)=angle(XYZ)CO interesect DO’ at FNote that (ECA)=(ABC)=betaAnd (ABD)=(ADE)=alphaQuadraliteral FCED is cyclic with (CFD)=(ECD)+(EDC)= alpha+beta=(CBD) (see picture)So 5 points F,C,E,D, B co-cyclicAnd ( EBD)=(ECD)= beta ( angles face the same arc DE)So x= (ABD)-(EBD)= alpha-betaPeter Tran
Join BC, BD∠CBD = ∠ABD + ∠ABC = α + β∠CED = 180° − (α + β)∴B, C, E, D are concyclicSo α = ∠CBE = x + ∠ABC = x + β∴ x = α − β
< CBA = beta and < ABD = alpha hence < CBD + CED = 180 So < EBD = beta = alpha - x hence x = alpha - betaSumith PeirisMoratuwaSri Lanka