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.
By construction, O is the orthocenter of ΔCDE, thus CO⊥DE. ∠BOA = 2∠C∠BDA = 180° − 2∠CDA = 2∠C (since DA=DC)Thus BDOA concyclic. Also, (since EB=EC)∠BEA = 180° − 2∠CThus BOAE concyclic. Hence, EAODB concyclic.
Problem 1113Is <MDC=<MDA=<ODA=90-<ACB=<AEO so EAOD is cyclic.Is <OAB=90-<OAB/2=90-<ACB=<ODC so AODB is cyclic.Therefore E,A,O,D and B are concyclic.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE