Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the diagram below to enlarge it.
Saturday, November 15, 2014
Geometry Problem 1060: Triangle, Incircle, Incenter, Inscribed circle, Tangent, Angle
Subscribe to:
Post Comments (Atom)
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