## Monday, May 19, 2008

### Elearn Geometry Problem 67

See complete Problem 67
Triangle, Circumcircle, Angles, Concyclic. Level: High School, SAT Prep, College geometry

1. Join B and E. Join A and E.
angle BAE = angle BED (angles subtended by equal arcs)
angle EAC = angle CBE (angles subtended by the same arc)
Hence angle BAE + angle EAC = angle BED + angle CBE
But angle BED + angle CBE = angle BGF (exterior angle of triangle GBE = sum of interior opposite angles)
Hence angle BAE + angle EAC = angle BGF
That is angle angle FAC = angle BGF
Hence A, F, G and C are concyclic.

2. Solution to Problem 67. In the circle circunscribed to ABC, ang(AFC) = (arc AC + arc BD)/2 and ang(AGC) = (arc AC + arc BE)/2. But arc BD = arc BE, so ang(AFC) = ang(AGC). These two angles subtend the same arc AC, thus AFGC is cyclic. In the circle circunscribed to AFGC, we have ang(CAG) = ang(CFG) because both subtend the same arc(CG).

3. Let PB be the tangent at B to the circumcircle
of ΔABC(such that P, D lie on the same side of AB).
Join BD, BE.
∠PBD
=∠BED (angle in the alternate segment)
=∠BDE (since BD = BE)
So PB ∥ DE.
Now ∠BFG
=∠PBF (alternate angles)
=∠PBA
=∠BCA (angle in the alternate segment)
Hence AFGC is a cyclic quadrilateral etc.