Geometry Problem

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

Click the figure below to see the complete problem 627.

## Wednesday, June 22, 2011

### Problem 627: Cyclic Quadrilateral, Congruence, Concyclic Points

Labels:
circle,
concyclic,
congruence,
cyclic quadrilateral

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Denote each of the equal angles ABD, ACD by x

ReplyDeleteAB = BE, CD=CF imply that

each of the angles AEB, CFD is (180° - x)/2

Taking supplements,

angle AED = angle AFD

So A,E,F,D are concyclic

The internal bisectors of the angles ABD and ACD

ReplyDeletemeet at the midpoint O of the (minor)arc AD.

Also O lies on the perpendicular bisector of AE as well as FD.

So OE = OA = OD = OF = k say

∴A, E, F, D lie on the circle (O;k)

Since < ABE = < ACD, Tr.s ABE & FCD are similar, hence < AED = < AFD and the result follows.

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka