Online Geometry theorems, problems, solutions, and related topics.
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.
Connect DE, ID and IENote that ID ⊥DC and IC ⊥ DE∠ (IFG)= ∠ (IDE) = alpha ( symmetric property of FBDG) But ∠ (IDE)= ∠ (DCI)= ∠ (ICE)= alphaSo ∠ (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. So alpha = < IEG = C/2Sumith PeirisMoratuwaSri Lanka