An acute-angled triangle PQR is inscribed in a circle with centre O.
S is the intersection of the bisector of P with QR
the perpendicular to PO through S meets the line PR in a point W interior to the segment PR.
***Source: A problem appearing in the Italian Mathematical Olympiad, 1995
Call the intersection of the perpendicular to PO through S the the bisector of angle RQP by T
Call angle RPQ by p
Call angle PQR by q
Construct segments OQ and OR.
by the given bisector angle RPS = angle QPS = p/2
by triangle PSQ angle PSQ = 180 - p/2 - q
by inscribed and central angle POR = 2q
by isosceles triangle POQ angle RPO = 90-q
by angle subtraction angle SOP = p/2+q-90
by right triangle STP angle TSP = 180-p/2-q
so angles PSQ and PSW are equal
by reflexive property SP=SP
because PS bisects angle RPQ angles RPQ and QPS are equal
so by ASA ΔPSW is congruent to ΔPSQ
so PW = PQ
Posted by Jer
on 2015-04-18 22:12:46