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

Labels:
angle,
circle,
circumcircle,
cyclic quadrilateral,
triangle

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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.