See complete Problem 77

Angles in a circle, Cyclic Quadrilateral, Parallel lines. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 77

Subscribe to:
Post Comments (Atom)

∠FEA=ext∠DCB=ext∠ABF

ReplyDeleteA,E,F,B concyc.

Join EF. AB is parallel to CD so angle BAE = angle CDE = angle CFE (angles in the same sector)

ReplyDeleteHence angle BAE = angle CFE. So AEFB is cyclic. Hence angle AEB = angle AFB (angles in the same sector)

Ajit: ajitathle@gmail.com

construction : join EF

ReplyDeleteangle DCF = angle DEF since they are in the same segment. similarly, we have angle CDE = angle CFD. ...(i)

but we are given that AB is parallel to CD, so angle ABC = angle BCD

angle BAD = angle ADC as they are alternate interior angles ...(ii)

but angle AEF + angle DEF = 180 degrees

and angle BFE + angle CFE = 180 degrees as they constitute linear pairs. ...(iii)

using the results of (i), (ii) and (iii) we get that angle ABF + angle AEF = 180 degrees

and angle BAE + angle BFE = 180 degrees

so that ABFE is a cyclic quadrilateral the sum of the two pairs of opposite angles being 180 degrees.

which implies that angle AEB = angle AFB.

Q. E. D.

< C = < E = < B

ReplyDeleteHence AEFB is cyclic and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

This picture can be considered the reciprocal of that of problem n° 72, where Reim’s direct therorem applies.

ReplyDeleteTherefore the reciprocal of Reim’s theorem can be applied here, and since AB//CD, therefore A, B, E, F belong to a circle O’ intersecting the given circle on E and F.

A, B, E, F being cocyclic, therefore <AEB=<AFB. q.e.d.

For a full development of Reim’s theorem, especially the many variety of configurations where it applies, read (in french): http://jl.ayme.pagesperso-orange.fr/apropos.html

Let the intersection of AD & BC be P

ReplyDeleteABFC is a cyclic quad.

=> alpha=beta (< in same seg.)