Dustan wrote: ↑Thu Dec 10, 2020 10:01 pm
Problem 51:Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
Solution :
Let $N$ be the intersection point of circumcircle of $\triangle BXM$ and $\triangle CYM$. $\angle BXM=\angle BNM$ and $\angle CYM=\angle CNM$. So, $MN$ bisects $\angle BNC$
$\therefore \frac{BN}{CN}=\frac{BM}{CM}=1\Rightarrow BN=CN$
So, $\triangle BNC$ is an isosceles triangle. Since $M$ is the midpoint of $BC$, we get that $NM\perp BC$. Also, $N,M,A$ collinear.
Now, $\angle PYN=\angle CYN=180^{\circ}-\angle CMN=90^{\circ}$
Similarly, $\angle PXN=90^{\circ}$.
So, $\angle PXN+\angle PYN=180^{\circ}$.
Therefore, $P,X,N,Y$ concyclic.
Again, since $BC||PA$, we get $\angle PAN=\angle BMN=90^{\circ}=\angle PYN$.
This implies $P,A,Y,N$ concyclic.
So, $P,A,X, Y, N$ concyclic. $Q.E.D.$