MHB Prove $\angle QBR=\angle RSQ$: Geometry Challenge

[anemone]

Let $A$ be the intersection point of the diagonals $PR$ and $QS$ of a convex quadrilateral $PQRS$. The bisector of angle $PRS$ hits the line $QP$ at $B$. If $AP\cdot AR+AP\cdot RS=AQ\cdot AS$, prove that $\angle QBR=\angle RSQ$.

 

[Albert1]

Let $A$ be the intersection point of the diagonals $PR$ and $QS$ of a convex quadrilateral $PQRS$. The bisector of angle $PRS$ hits the line $QP$ at $B$. If $AP\cdot AR+AP\cdot RS=AQ\cdot AS$, prove that $\angle QBR=\angle RSQ$.

Spoiler

View attachment 2866

 

[anemone]

Thanks for participating, Albert and thanks for your diagram and solution!

Here is a quite similar approach that I saw that I wanted to share with the community of MHB, since it uses one theorem that I have never heard before, that is, the "Power of a Point Theorem", for those who have never heard of this theorem before, I hope you will like it "at first sight" as much as I do:

Spoiler

View attachment 2885

Let $N$ be the intersection of lines $BR$ and $QS$. By the angle bisector theorem applied to triangle $ARS$, we have $\dfrac{RS}{SN}=\dfrac{AR}{AN}\,\,\rightarrow\,\,RS=\dfrac{AR\cdot SN}{AN}$.

Substitute this into the given relation, we then have

$\begin{align*}AQ\cdot RS&=AP\cdot AR+AP\cdot RS\\&=AP\cdot AR+\dfrac{AP\cdot AR\cdot SN}{AN}\\&=AP\cdot AR\left(1+\dfrac{SN}{AN}\right)\\&=AP\cdot AR \cdot \dfrac{AS}{AN}\end{align*}$

Simplify this gives $AQ\cdot AN=AP\cdot AR$.

Because $A$ lies inside quadrilateral $PQRN$, the Power of a Point theorem implies $P,\,Q,\,R,\,N$ are concyclic. Hence $\alpha=\angle BQS=\angle PQN\stackrel{\small \text{angle subtended on the same arc}}{=}\angle PRN\stackrel{\small \text{angle bisector theorem}}{=}\angle NRS=\angle BRS$. This implies that $B,\,Q,\,R,\,S$ are concyclic. Therefore $\angle QBR=\angle RSQ$.