Q3 from British Mathematical Olympiad Paper 2 1996 I have been trying to practice my maths by going through old BMO papers and have come across this question for which I have no idea. They do not display answers for questions this old anymore.
Distinct points A, P, Q, R, and S  lie on the circumference of a circle and AP, AQ, AR, and AS are
chords with the property that:
∠PAQ = ∠QAR = ∠RAS.
Prove that
AR(AP + AR) = AQ (AQ + AS).
SOURCE:
https://bmos.ukmt.org.uk/home/bmolot.pdf Page 4
NOTE: If needed could people edit the tags because I am not fully sure what tags are most appropriate. 
 A: 
$$\begin{align}
\angle PAQ=\angle QAR=\angle RAS &\quad\implies\quad |PQ|=|QR|=|RS| &=: a \\
\angle PAR=\angle QAS &\quad\implies\quad |PR|=|QS| &=: b
\end{align}$$
Now, we simply invoke Ptolemy's Theorem on two inscribed quadrilaterals:
$$\left.\begin{align}
\square APQR: &\quad p a + r a = q b \\
\square AQRS: &\quad q a + s a = r b
\end{align}\quad\right\}\quad\implies\quad \frac{p+r}{q} = \frac{b}{a} = \frac{q+s}{r}\quad\implies\quad r(p+r) = q(q+s)$$
A: A synthetic solution:
The basic principle here? Keep adding things to the picture until we have enough to make it work.
First, the basic diagram with our five points $A,P,Q,R,S$ on a circle $\Gamma$ and some equal angles marked. Also, for future use, I've labeled the segments $AQ$ and $AR$.

Second, that expression $AR(AP+AR)$ kind of reminds me of Power of a Point. Let's make it actually into a case of that - extend $PA$ to a point $B$ such that $AB=AR$, and extend $SA$ to $C$ such that $AC=AQ$. Also, draw in $BR$ and $CQ$ to close those isosceles triangles:

Well, now we have two different points that we're computing powers from. It's the same circle $\Gamma$, so we win if they're the same distance from the center. That's not much use directly, but the new lines we've drawn in are interesting. The isosceles triangles make $\angle AQC =\angle QCA=\frac12\angle QAS = \angle QAR$, so $CQ$ and $AR$ are parallel. Similarly, $BR$ and $AQ$ are parallel. Extend $BR$ and $CQ$ to meet at $D$ and complete the parallelogram:

We have $\angle CDB = \angle ACD = \angle DBA$, so $ACDR$ and $AQDB$ are isosceles trapezoids. The perpendicular bisector of $AQ$, which passes through the center of $\Gamma$, also bisects $BD$, so $D$ is the reflection of $B$ across this bisector. In particular, if $R'$ is the second intersection of $BD$ with $\Gamma$, then $BR'=DR$ and $BR=DR'$, so $DR\cdot DR'=BR'\cdot BR=BA\cdot BP=AR\cdot (AR+AP)$. Similarly, let $Q'$ be the second intersection of $CD$ with $\Gamma$. $D$ is the reflection of $C$ across the perpendicular bisector of $AR$, which also passes through the center of $\Gamma$ and bisects $CD$, so $CQ'=DQ$ and $CQ=DQ'$. Thus $DQ\cdot DQ' = CQ'\cdot CQ=CA\cdot CS = AQ\cdot (AQ+AS)$. But of course, $DQ\cdot DQ'=DR\cdot DR'$, so $AR\cdot (AR+AP)=AQ\cdot (AQ+AS)$. Done.
As it happens, $QQ'=AP$ and $RR'=AS$. We didn't use these directly, but in the end we can mark off an awful lot of equal lengths in the picture. My method is certainly not the only one that works well.
