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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1209.
Let F is the point of tangent of BC to circle ESince E is center of excircle of triangle ABC => ∠ (ACE)= ∠ (ECF)∠ (ACE) subtend arc DE∠ (ECF) subtend arc EC+ arc BC= arc BESince ∠ (ACE)= ∠ (ECF)=> arc DE= arc BE=> DE=BE
Is <CBE=<DCE=<ECF (EF⊥BC).But <ECF=<BDE.Therefore <DBC=<BDE or BE=DE.
Draw DM tg to E, N tg point of BC =>BE bisector, DE bisector =>R tr EDM congr ENB
< BEC = 90 - C/2 - B/2 = A/2 = < BDC = < DBA since < BAC = A Further easily < EDC = < EBC = B/2 and so < BDE = A/2 + B/2 = < DBE and so BE = DESumith PeirisMoratuwaSri Lanka
Further AD = AB as well
Much simpler proof.....< DBE = < DCE = < BDE and the result follows
Let F be the point of tangent of BC to circle E on line BC∠DBE=∠DCE=∠ECF=∠BDE∴BE=DE