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

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

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## Saturday, October 1, 2016

### Geometry Problem 1269 Triangle, Incircle, Incenter, Inscribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector

Labels:
angle bisector,
circle,
incenter,
incircle,
inscribed,
perpendicular,
tangency point,
tangent,
triangle

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connect OC, <BDA1 = <COA1, so OCDA1 is on the same circle. <ODC = 90 = <OA1C =<AA1C

ReplyDelete< DEB = 90-B/2 and < A1AE = A/2 so

ReplyDelete< EA1A = 90-B/2 - A/2 = C/2.

But < OCD = C/2 hence OCDA1 is concyclic

Therefore <OA1C = <ODC = 90

Sumith Peiris

Moratuwa

Sri Lanka

It follows that < OCA1 = B/2

ReplyDelete