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.)


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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Solution

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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.)


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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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 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 © 1996-2018 Alexander Bogomolny

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