Let PQRS be a parallelogram and T be a point on diagonal QS such that ∠TRQ = ∠ PRS.
The circumcircle of triangle PQS intersects line PR at points P and U.
Find : ∠ PUS/∠ TUQ.

** Source: Serbia National Math Olympiad.

Let a line from the centre, O, of the circumcircle, through the mid-
point of chord SQ cross PQ at V. Let SV and SR cross the circle
again at W and X respectively, and let WU cross SQ at Y.

arc SP = arc WQ             (symmetry about line OV)
arc SP = arc QX  (arcs between parallel chords)
/WSQ = /WUQ = /PUS = /QSX = /PQS = a say   (angles in same seg.)
and       /PSW = /PUW = /PQW = b say  (angles in same segment)

Also      /SQR = /QSP = a + b      (Alternate angles in parallelogram)

Therefore, in triangles WQY and RQT:
            /WQY = /RQT     (both equal to a + b)
            WQ = RQ           (both equal to SP)
            /QWY = /QRT     (/QWY = /QPU angles in same segment,
                                     /QPU = /PRS are alternate & /PRS = /QRT)

Thus, triangles WQY and RQT are congruent, and the points T and Y
coincide.                       
So  /TUQ = /WUQ = a = /PUS,     giving   /PUS / /TUQ = 1

Edited on October 6, 2015, 5:08 pm

Posted by Harry on 2015-10-06 17:07:10