Problem:
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.
Solution:
First, we prove the "if" part by assuming that ray BG bisects ∠CBD; that is, we assume that DQ=CQ.
It is easy to see that ABCD is an isosceles trapezoid with AD=BC. In particular, AD=BC and AC=BD.
Because ABCPD is cyclic, it follows that
∠APC=2AC=2BD=∠BCD=∠SCD and ∠APD=2AD=2BC=∠BDC=∠RDC.
Because RS∥DC, it follows that 180∘=∠GRD+∠RDC=∠GRD+∠APD and 180∘=∠GSC+∠SCD=∠GSC+∠APC; that is, both GSCP and GRDP are cyclic. Hence, ∠GPR=∠GDR and ∠GPS=∠GCS. In particular, we have
∠RPS=∠GPR+∠GPS=∠GDR+∠GCS(3)
Let K be the intersection of segments BQ and CD. We have ∠CBK=∠QBD and ∠KCB=∠DCB=∠DQB; that is, triangles CBK and QBD are similar to each\
other. Because RG∥CD, we have BG/GK=BR/RD. This means that G and R are the corresponding points in the similar triangles CBK and QBD. Consequently, we have ∠BCG=∠BQR. In exactly the same way, we can show that ∠BDG=∠BQS. Combining the last two equations together with (3) yields
∠RQS=∠BQS+∠BQR=∠BDG+∠BCG=∠RDG+∠SCG=∠RPS
from which it follows that PQRS is cyclic.
Second, we prove the "only if" part by assuming that PQRS is cyclic. Let γ denote the circumcircle of PQRS. We approach indirectly by assuming that ray BG does not bisect ∠CBD. Let G1 be the point on segment RS such that ray BG1 bisects ∠CBD. Let rays AG1 and BG1 meet ω again at P1 and Q1 (other than A and B ). By our proof of the "if" part, P1Q1RS is cyclic, and let γ1 denote its circumcircle.
Hence lines RS,PQ,P1Q1 are the radical axes of pairs of circles γ and γ1,γ and ω,γ1 and ω, respectively. Because segments P1 is the midpoint of arc CD (not including A and B), lines P1Q1⊮CD, implying that lines P1Q1 and RS intersect, and let X denote this intersection. Thus X is the radical center of ω,γ,γ1. In particular, line PQ also passes through X. We obtain the following configuration.
There are two possibilities for the position of line PQ, namely, (1) both P and Q lie on minor arc P1Q1; (2) one of P and Q lies on minor arc DQ1 and the other lies on minor arcP1B. If G lies on segment RG1, then Q lies on minor arc DQ, and we must have (2). But in this case, P must lie on minor arc Q1P1, violating (2). If G lies on segment G1S, then P must lie on minor arc P1B, and again we must have (2). But in this case, Q must lie on minor arc Q1C, violating (2). In every case, we have a contradiction. Hence our assumption was wrong, and ray BG bisects ∠CBD.
Solution 2. We present another approach of the "if" part.
Let rays CG and DG meet ω again at E and F, respectively. Let R1 denote the intersection of segments BD and QE, and let S1 denote the intersection of segments BC and QF. Applying Pascal's theorem to cyclic hexagon BDFQEC shows that R1,G,S1 are collinear. Because
∠R1EG=∠QEC=2CQ=2DQ=∠DBQ=∠R1BG
we deduce that EBGR1 is cyclic. Because EBGR1 and EBCD are cyclic, we have
∠BR1S1=∠BR1G=∠BEG=∠BEC=∠BDC
from which it follows that R1S1∥CD; that is, R1=R and S1=S.
Therefore, (3) becomes
∠RPS=∠GDR+∠GCS=∠FDB+∠BCE=∠FQB+∠BQE=∠FQE=∠RQS, implying that PQRS is cyclic.
