Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, April 16, 2016

### Geometry Problem 1209: Triangle, Circle, Excircle, Excenter, Circumcircle, Congruence

Labels:
circle,
circumcircle,
congruence,
excenter,
excircle,
triangle

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Let F is the point of tangent of BC to circle E

ReplyDeleteSince E is center of excircle of triangle ABC => ∠ (ACE)= ∠ (ECF)

∠ (ACE) subtend arc DE

∠ (ECF) subtend arc EC+ arc BC= arc BE

Since ∠ (ACE)= ∠ (ECF)=> arc DE= arc BE=> DE=BE

Is <CBE=<DCE=<ECF (EF⊥BC).But <ECF=<BDE.Therefore <DBC=<BDE or BE=DE.

ReplyDeleteDraw DM tg to E, N tg point of BC =>

ReplyDeleteBE bisector, DE bisector =>

R tr EDM congr ENB

< BEC = 90 - C/2 - B/2 = A/2 = < BDC = < DBA since < BAC = A

ReplyDeleteFurther easily < EDC = < EBC = B/2 and so < BDE = A/2 + B/2 = < DBE and so BE = DE

Sumith Peiris

Moratuwa

Sri Lanka

Further AD = AB as well

ReplyDeleteMuch simpler proof.....

ReplyDelete< DBE = < DCE = < BDE and the result follows

Let F be the point of tangent of BC to circle E on line BC

ReplyDelete∠DBE=∠DCE=∠ECF=∠BDE

∴BE=DE