## Saturday, April 16, 2016

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

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. 1. Let F is the point of tangent of BC to circle E
Since 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

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

3. Draw DM tg to E, N tg point of BC =>
BE bisector, DE bisector =>
R tr EDM congr ENB

4. < 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 = DE

Sumith Peiris
Moratuwa
Sri Lanka

5. Further AD = AB as well

6. Much simpler proof.....

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

7. Let F be the point of tangent of BC to circle E on line BC
∠DBE=∠DCE=∠ECF=∠BDE
∴BE=DE