$AP, AQ, AR, AS$ are chords of a given circle with the property that $\angle PAQ = \angle QAR = \angle RAS$. Prove that $AR(AP+AR) = AQ(AQ+AS)$.
This was question 3 of the 1994 BMO2 (https://bmos.ukmt.org.uk/home/bmo2-1994.pdf). I can't seem to get a grip on the problem. Can anyone help me out by providing a starting point?
First of all consider $PAQ$ and $QAR$. We could get that : $AQ^2 + AR^2 -2AQ \cdot AR\cdot cos\alpha = AP^2 + AQ^2 -2AP \cdot AQ \cdot cos\alpha$, so $cos\alpha = \frac{AP + AR}{2AQ}$. And find the same angle in $RAS$ and $RAQ$ triangles we will have : $\frac{AP + AR}{2AQ} = \frac{AS + AQ}{2AR}$. That give us $AR(AR + AP) = AQ(AQ + AS)$