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 20151006 17:07:10 