See complete Problem 72

Intersecting Circles, Cyclic Quadrilateral, Angles, Parallel. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 72

Labels:
angle,
circle,
cyclic quadrilateral,
parallel

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ABCD is cyclic so angle BAD = angle DCF. Similarly DCFE is cyclic and angle DCF = 180 -angle FED. Hence AB is parallel to EF.

ReplyDeleteWhat about when BF intersects AE? that is another case, easy to proof, but is part of the problem.

DeleteGreetings

Your case is just one of many varieties of configurations where Reim’s theorem can be applied straightforwardly.

DeleteEnjoy!

This configuration is the general case illustrating Reim’s theorem.

ReplyDeleteAnton Reim (1832–1922), czech mathematician.

This theorem can easily be applied to some of the next problems proposed by Antonio Gutierrez: n° 73, 74, 75.

The reciprocal il also true and is illustrated by the configuration shown in pb n° 77, where CD is a given chord in circle O, lines DEA and CFB with A and B such as AB//CD. The reciprocal of Reim’s theorem states that A,B,E,F are concyclic in circle O’ which intersects circle O in E and F.