# 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? |

|Activities| |Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2017 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 © 1996-2017 Alexander Bogomolny

62686554 |