See complete Problem 67
Triangle, Circumcircle, Angles, Concyclic. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 67
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Join B and E. Join A and E.
ReplyDeleteangle 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 quadrilateral FACG is cyclic.
Hence A, F, G and C are concyclic.
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).
ReplyDeleteLet PB be the tangent at B to the circumcircle
ReplyDeleteof 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.
Draw BD and BE; BD=BE
ReplyDeleteang(BDE)=ang(DEB)=a ; ang(EBG)=b => ang(FGB)=a+b => ang (FGC)=180-a-b
ang(FAC)=angBAC)=ang(BDE)+ang(EBC)=a+b => ACGF is concyclic
ang(GFC) and ang(GAC) on the same arc. => ang(GFC)=ang(GAC)
Let BT be the tangent in B of the circumcircle of ABC.
ReplyDeleteB is the center of arc DE, therefore BT//DE
Then, applying Reim’s reciprocal theorem, A,C,G,F are concyclic.
(For this theorem, see my solution of problem n°77
For the direct Reim’s theorem, see my solution of problem n°72)