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 1270 Triangle, Excircle, Excenter, Escribed Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector

Labels:
angle bisector,
escribed circle,
excenter,
excircle,
perpendicular,
tangency point,
tangent,
triangle

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< EAO = < EC1O = 90-A/2

ReplyDelete(since < C1CD = 90-C/2 and < C1DC = 90-B/2)

Hence EAC1O is a cyclic quadrilateral

So < AC1C = < AEO = 90

Sumith Peiris

Moratuwa

Sri Lanka

Further if AO meets DE at A1 and CA1 and C1A meet at X then O,X,F are collinear points.

ReplyDeleteFurther if CX cuts FD at Z and AX cuts FE at Y then XF and YZ bisect each other. (XYFZ is a parallelogram)