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

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

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## Saturday, November 15, 2014

### Geometry Problem 1060: Triangle, Incircle, Incenter, Inscribed circle, Tangent, Angle

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Connect DE, ID and IE

ReplyDeleteNote that ID ⊥DC and IC ⊥ DE

∠ (IFG)= ∠ (IDE) = alpha ( symmetric property of FBDG)

But ∠ (IDE)= ∠ (DCI)= ∠ (ICE)= alpha

So ∠ (DCE)= 2 alpha

IDCE is cyclic so < IDE = C/2. Hence < BGD must be A/2 considering the angles of Tr. BDG = < IAE. Hence AIGE is cyclic on which circle F must also lie since AEIF is obviously cyclic.

ReplyDeleteSo alpha = < IEG = C/2

Sumith Peiris

Moratuwa

Sri Lanka