## Monday, May 19, 2008

### Elearn Geometry Problem 77

See complete Problem 77
Angles in a circle, Cyclic Quadrilateral, Parallel lines. Level: High School, SAT Prep, College geometry

1. ∠FEA=ext∠DCB=ext∠ABF
A,E,F,B concyc.

2. Join EF. AB is parallel to CD so angle BAE = angle CDE = angle CFE (angles in the same sector)
Hence angle BAE = angle CFE. So AEFB is cyclic. Hence angle AEB = angle AFB (angles in the same sector)
Ajit: ajitathle@gmail.com

3. construction : join EF
angle 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.

4. < C = < E = < B

Hence AEFB is cyclic and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

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

6. Let the intersection of AD & BC be P