If $AB = AC$, the statement is obvious. In the following, we assume $AB < AC$, the other case being similar.

* Triangles $PAM$ and $PBM$ are congruent (SSS), hence $\angle PMA \equiv \angle PMB$. Similarly, $\angle QMA \equiv \angle QMC$, and this leads rapidly to $\angle PMQ = 90^\circ$.

As $\tan B = \frac{AH}{BH} = \frac{AC}{AB} = \frac{AQ}{BP}$ and $\angle HAQ = \angle HBP = 60^\circ + \angle B$, triangles $HAQ$ and $HBP$ are similar (SAS).

It follows that $\angle QHA = \angle PHB$, hence $\angle QHP = \angle QHA + \angle AHP = \angle PHB + \angle AHP = \angle AHB = 90^\circ$.

In conclusion, the points $P, Q, M$ and $H$ are co-cyclic, hence $\angle HPQ \equiv \angle CMQ \equiv \angle AMQ$.

Let $\{D\} = HQ \cap MP$. The angle $\angle MNQ$ is exterior to the triangle $PMN$, therefore $\angle MNQ = \angle MPN + \angle NMP = \angle MHQ + \angle BMP = \angle MDQ$. (1)

It follows that the quadrilateral $NDMQ$ is cyclic, hence $\angle DNQ = 90^\circ$. This means that $NDHP$ is also cyclic, therefore $\angle PNH \equiv \angle PDH$. (2)

From (1) and (2) it follows that $\angle PNH \equiv \angle PDH \equiv \angle MDQ \equiv \angle MNQ$, i.e. ($ND$ is the bisector of angle $\angle HNM$). For the triangle $HMN$ ray ($NP$ is the external bisector, while ($MP$ is an internal bisector, which means that $P$ is the excenter opposite to the vertex $M$). Then ($HP$ is the internal bisector of angle $\angle NHB$.

Finally, $\angle NHP \equiv \angle PHB \equiv \angle AHQ$.

(given in the contest by Paul Bécsi)

Let $\{S\} = AH \cap PQ$. An easy computation shows that $\angle SAQ = \angle NAP = 120^\circ - \angle B$, which shows that the rays ($AN$ and ($AS$ are isogonal in the angle $\angle PAQ$). From Steiner's Theorem it follows that $\frac{PN}{NQ} \cdot \frac{PS}{SQ} = \left(\frac{PA}{AQ}\right)^2$.

The conclusion means the rays ($HN$ and ($HS$ are isogonal in the angle $\angle PHQ$, which, according to Steiner's Theorem, is equivalent to $\frac{PN}{NQ} \cdot \frac{PS}{SQ} = \left(\frac{PH}{HQ}\right)^2$.

In conclusion, we need to prove that $\frac{PA}{AQ} = \frac{PH}{HQ}$.

But $\Delta BHA \sim \Delta BAC$ leads to $\frac{BH}{HA} = \frac{BA}{CA} = \frac{BP}{CQ}$. As $\angle PBH = 60^\circ + \angle B = \angle QAH$, it follows that $\Delta PBH \sim \Delta QAH$, i.e., $\frac{PH}{QH} = \frac{PB}{AQ} = \frac{PA}{AQ}$ and the conclusion.