Angle Bisector in Equilateral Trapezoid
The applet below illustrates problem 5 from the 2009 USA Mathematical Olympiad:
Trapezoid ABCD, with AB||CD, is inscribed in circle ω and point G lies inside triangle BCD. Rays AG and BG meet ω again at points P and Q, respectively. Let the line through G parallel to AB intersect BD and BC at points R and S, respectively. Prove that quadrilateral PQRS is cyclic if and only if BG bisects ∠CBD.
(Note that, in the applet, the effect of dragging points A, B, C, D depends on which of the two boxes - "Adjust circle" or "Move points" - is checked.)
What if applet does not run? |
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Copyright © 1996-2018 Alexander Bogomolny
Trapezoid ABCD, with AB||CD, is inscribed in circle ω and point G lies inside triangle BCD. Rays AG and BG meet ω again at points P and Q, respectively. Let the line through G parallel to AB intersect BD and BC at points R and S, respectively. Prove that quadrilateral PQRS is cyclic if and only if BG bisects ∠CBD.
(Note that, in the applet, the effect of dragging points A, B, C, D depends on which of the two boxes - "Adjust circle" or "Move points" - is checked.)
What if applet does not run? |
Several solutions have been posted at the mathlinks forum. The one below is, in my view, the most straightforward.
Assume ∠CBQ = ∠QBD. Let BQ, DC intersect at T. Obviously triangles DTB and QCB are similar. This gives DGB and QSB similar since
Assume PQRS is cyclic. It is easy to know ∠AQR = ∠BPS since both ABPQ and RSPQ are cyclic.Also,
|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|
Copyright © 1996-2018 Alexander Bogomolny
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