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.)

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This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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Solution

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Copyright © 1996-2012 Alexander Bogomolny

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 BG/BT = BS/BC. So ∠BDG = ∠BQS. Since ∠GPD = ∠BDC = ∠BRS so DRGP is cyclic. Further, ∠BDG = ∠RPG. So ∠BQS = ∠RPG and, similarly, ∠BQR = ∠SPG. Adding the two gives ∠RQS = ∠RPS. So PQRS cyclic.

Assume PQRS is cyclic. It is easy to know ∠AQR = ∠BPS since both ABPQ and RSPQ are cyclic.Also, ∠AQB = ∠APB, implies ∠BQR = ∠SPG. Then ∠SPQ = ∠SCG since GSPC is cyclic (because ∠GSC = ∠GPC). Tthen ∠BQR = ∠BCG. Now suppose ∠DBQ bigger than ∠∠CBQ. Then ∠BRQ is smaller than ∠BGC. Since they are both bigger than 90°, sin(BRQ) > sin(BGC). Using the Sine Law in triangles BRQ and BGC and simplifying you can get BQ/BC > BD/BM. Now by proving ∠BCQ > ∠BMD > 180° - ∠BCQ, you know that sin(BMD) > sin(BCQ). From the Sine Law in triangles BDM and BCQ you can get BQ/BC < BD/BM, which is a contradiction. The other way is similar (but not completely same though, the idea is the same, just using sine law). Hence ∠DBQ cannot be bigger than or smaller than ∠CBQ.

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Copyright © 1996-2012 Alexander Bogomolny