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

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

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## Monday, April 27, 2015

### Geometry Problem 1113: Triangle, Perpendicular Bisector, 90 Degrees, Circle, Concyclic Points, Cyclic Quadrilateral

Labels:
90,
circle,
concyclic,
cyclic quadrilateral,
perpendicular bisector,
triangle

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By construction, O is the orthocenter of ΔCDE, thus CO⊥DE.

ReplyDelete∠BOA = 2∠C

∠BDA = 180° − 2∠CDA = 2∠C (since DA=DC)

Thus BDOA concyclic.

Also, (since EB=EC)

∠BEA = 180° − 2∠C

Thus BOAE concyclic.

Hence, EAODB concyclic.

Problem 1113

ReplyDeleteIs <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