Let $ABC$ be an acute angled scalene triangle. Let $ABC$ be an acute angled scalene triangle. Let $P$ be a point on the extension of $AB$ past $B$, and $Q$ a point on the extension of $AC$ past $C$ such that $BPQC$ is a cyclic quadrilateral. Let $N$ be the foot of the perpendicular from $A$ to $BC$. If $NP = NQ$ then prove that $N$ is also the centre of the circumcircle of $APQ$.
 A: Since $BCPQ$ is a cyclic quadrilateral, the $BC$-line is "antiparallel" to the PQ-line. Since in a general triangle the orthocenter $H$ and the circumcenter $O$ are isogonal conjugates, the $AN$-line goes through the orthocenter of $ABC$ and the circumcenter of $APQ$. Then, provided that $N$ lies on the perpendicular bisector of $PQ$, $N\equiv O$ is the circumcenter of $APQ$.

A: Let us denote the angles of $\triangle ABC$ by $\alpha, \beta, \gamma$. Note that $BPQC$ being a cyclic quadrilateral is equivalent to $\angle AQP = \angle CBA = \beta$. With $O$ being the circumcenter of $\triangle APQ$ we have for the center angle above $PQ$
$$\angle POQ = 2 \angle PAQ = 2\alpha.$$ 
Hence in isosceles $\triangle OPQ$ we get
$$\angle OPQ = \angle OQP = (180 - 2\alpha)/2 = 90-\alpha.$$
Now we can compute
$$\angle AQO = \angle AQP - \angle OQP = \beta - (90 - \alpha) = 90 - \gamma.$$
In isosceles $\triangle AOQ$ we have
$$\angle OAQ = \angle AQO = 90 - \gamma,$$
which is the same as $\angle NAC$. Hence $N$ is on $OA$. Since $O$ and $N$ are both on the perpendicular bisector of $PQ$ and $\triangle APQ$ is scalene (as $\triangle ABC$ is), $O$ and $N$ must coincide. q.e.d.
A: Let us write $\angle CAB=\beta,\angle ACB=\gamma,$ and $\angle NPQ=\theta=\angle NQP$ (because triangle $PNQ$ is isosceles). Since $BPQC$ is cyclic, $\angle APQ=\angle ACB=\gamma$, and so $\angle NPA=\gamma-\theta$. Also, $\angle NAP=\angle NAB=\frac12\!\pi-\beta$ since $NA\perp NB$. Similarly, $\angle NQA=\beta-\theta$ and $\angle NAC= \frac12\!\pi-\gamma$. Applying the sine rule  to triangles $NPA$ and $NPQ$ gives $NP/NA=\sin(\frac12\!\pi-\beta)/\sin(\gamma-\theta)$ and $NQ/NA=\sin(\frac12\!\pi-\gamma)/\sin(\beta-\theta)$ respectively. These ratios are equal, and so $$\sin(\beta-\theta)\cos\beta=\sin(\gamma-\theta)\cos\gamma.$$This may be written $\sin(2\beta-\theta)-\sin\theta=\sin(2\gamma-\theta)-\sin\theta$ or$$\sin(2\beta-\theta)=\sin(2\gamma-\theta).$$Since $\beta\neq\gamma$, we have $$2\beta-\theta=\pi-(2\gamma-\theta),$$ or $\theta=\beta+\gamma-\frac12\!\pi$. Then $\angle NPA=\gamma-\theta=\frac12\!\pi-\beta=\angle NAP$. It follows that $NA=NP=NQ$, and hence $N$ is the circumcentre of triangle $APQ$.
A: 
The red circle ($\pi$) circum-scribes APQ. Produce AN to cut ($\pi$) at G.
$\gamma = \delta$ .... [ext. angles cyclic quad.]
$= \epsilon$ .... [angles in the same segment]
This means, $BNGP$ is cyclic. This further means, $\alpha = \beta = 90^0$
Therefore, ANG is the diameter of $\pi \tag 1$
Let $H$ be the midpoint of $PC$ [EDIT:- $PQ$ instead]. Then, $\triangle NHP \cong\triangle NHQ$.
Therefore, $NH$ is the perpendicular bisector of the chord $PQ \tag 2$
$N$ is the intersection of (1) and (2) implies N is the center of $\pi$.  
