Clearly the problem is actually just triangle configuration geometry: Let \(\triangle ABC\) have \(A\)-Dumpty point \(M\), and let \(P\) and \(Q\) be the points on \((ABC)\) (on minor arcs \(AB\) and \(AC\) respectively) such that \(\angle PMB = \angle QMC = 90^{\circ}\). Then we want to show that \(AM\), \(BP\), \(CQ\) concur. Now note that if \(K\) is the other intersection of \(AM\) and \((ABC)\), then for the problem to be true, we must have \[-1 = (B, C; A, K) = (P, Q; A, K)\] where we project through \(X = BP \cap QC\) assuming it lies on \(AK\). Furthermore, let \(Y = PC \cap AM\), and projecting \((P, Q; A, K)\) through \(Y\), we get \[-1 = (C, QY \cap (ABC); K, A) \implies Q - Y - B\] so \(PC\), \(BQ\), \(AM\) must concur (again, if the problem is true). But here's the fun part - due to radical axis, we now must have five points on \(AM\) such that \(\angle PZB = \angle QZC\) - namely, \(X\), \(Y\), \(M\), and \((PBM) \cap (QCM)\), \((PBY) \cap (QCY)\). These points being all distinct is not possible, and so \(Y\) must be the second intersection of \((PBM)\) and \((QCM)\) (if the problem is true).

But this means that what we have to show is just that \(M\) is the spiral center sending \(BP \to QC\). This finishes because it follows that \(Y = BQ \cap CP = (BPM) \cap (QCM)\), and after a short chase looking at \(\angle AMC\), we get \(Y \in AM\), and we are done through radical axis. If \(M\) is the spiral center \(BP \to QC\), then \(MB \cdot MC = MP \cdot MQ\) must be true, which means \(MA^2 = MP \cdot MQ\) must be true. However, \(M\) is the midpoint of the symmedian chord, so \(MA \cdot MK = MP \cdot MQ\) is wanted, which means that if \(PM \cap (ABC) = Q'\) and \(QM \cdot (ABC) = P'\), then \(MP = MP'\) and \(MQ = MQ'\) must hold. But of course this is true since \(AM\) is the angle bisector of \(\angle PMQ\) and \(OM \perp AM\).

Indeed, note that all we want to show is \((A, K; P, Q) = -1\). But using lemma 9.11, we know \((MA, MO; MP, MQ) = -1\), so all we want to show is that \(T = KK \cap BC = AA \cap BC\) lies on \(OM\), which is well-known (radical axis on \((AO)\), \((BOC)\), \((ABC)\)).