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.
Let DO & CI, both extended, meet in X. /_AOC=2B since O is the circumcentre and thus /_COI=/_B. Since DI//BA, /_CDI=/_B. Quad CDOI is, therefore, concyclic and thus, /_IOX=/_DCI=/_C/2. We note that /_OIX=90° - /_C/2 which, in turn, means that /_DXI =180° - (90-C/2) - C/2 = 90°. QED.
"Quad CDOI is, therefore, concyclic"How does it follow from your previous statements?
< IOC = < B from Tr. BOC = < IDC since ID //AB. So IODC is cyclic and so < MOI = < ICD = < A/2 = < ICN. Hence MOCN is cyclic and so < OMC = < ONC = 90.Note - M is where DO meets CI and N is the mid point of AC (note that B, O, I, N are all collinear since Tr. ABC is isoceles) Sumith PeirisMoratuwaSri Lanka