Question: Let P, Q, R be points on the sides AB, BC, CA, respectively,
of a triangle ABC. Assume that the circumscribed circles of the triangles
PBQ and QCR intersect at points Q and S. Prove that the points A, P, S, R
lie on a common circle
Here is what I have so far:
The points APSR form a quadrilateral. The points of this quadrilateral all lie on a common circle if and only if its opposite angles add up to 180 degrees.
To prove they all lie on a common circle we can look at the circumscribed circle of 3 of the points and try to prove somehow that the 4th point lies on the same circle. The circle that seems best for this is the circumscribed circle of ARP. I am not sure how to go about this.
Maybe one can look at the angles subtended by various arcs on the circle and come to a conclusion somehow? For example angle ASP is equal to angle ARP if the points lie on the same circle as the angles are subtended by the same arc.
Any help is appreciated.
Best Answer
You were in the right direction: The opposite angles of a cyclic quadrilateral add up to $180$. So: $$\angle PBQ + \angle PSQ = 180 \text{ and } \angle RCQ + \angle RSQ = 180.$$ That is $$ \angle PBQ + \angle PSQ + \angle RCQ + \angle RSQ = 360,$$ and so $$ (\angle PBQ + \angle RCQ + \angle PAR) + (\angle PCQ + \angle RSQ + \angle PSR) = 360 + \angle PAR + \angle PSR.$$ Now note that in the above identity, $$\angle PBQ + \angle RCQ + \angle PAR = 180 \text{ and } \angle RSQ + \angle RSQ + \angle PSR = 360.$$