Solution: a) Simson Lines $R, Q, W$ and $Q, W, S$, done. b) Trig bash with Ptolemy. $\blacksquare$

My approach for b. We get that $RS - QW = RW + QS$ and $\frac{RW}{MP} = \frac{CW}{CP} = \frac{CN}{BC}$ also $\frac{QS}{MP} = \frac{BQ}{BM} = \frac{BN}{BC}$. Thus $\frac{RW}{MP} + \frac{QS}{MP} = \frac{CN}{BC} + \frac{BN}{BC} = 1$. Hence the result.

a)True by simson-line theorem b)We have $ RS-QW=(AN-HN) \cdot sinA=AH \cdot sinA $.

Dear Mathlinkers, we can get the collinearity by involving the Miquel's line of A' wrt the complete quadrilateral AMPH. Sincerely Jean-Louis

For part a) - Notice that $N\in (AMB)$ so from Simpson we obtain that $S, Q, R$ are collinear. Similarly, $N\in (APC) \Rightarrow S,W,R$ are collinear. Hence, $S,W,Q,R$ are collinear. For part b) - Clearly, $MP||SR$ as $\angle PMB=\angle RSN$ and $NS||MB$. Define point $L$ such that $PL||WQ$ and $WQLP$ is a parallelogram. Therefore, $M,P,L$ are collinear, $WP=QL$ and $WP||QL (1)$. It is enough to show that $MLRS$ is also a parallelogram. Clearly, $WP=NR$ and $WP||NR (2)$ . Combining $(1)$ and $(2)$, we get that $LRNQ$ is also a parallelogram. Hence, $LR=QN$ and $LR||QN$. However, because $MSNQ$ is a rectangle, we get that $SM=QN$ and $MS||QN$. So $MS=LR$ and $MS||LR$ and we are done.