It is suffice to show $OP=OQ$ We are going to use complex numbers. Consider $(ABC)$ as the unit circle centered at origin $O$ in the complex plane and small letter of vertices denote vertices in complex plane We know $O=0$ $h=a+b+c$ $m=\frac{2a+b+c}{2}$ Let $X$ be a point such that $XM \perp OM$. So $\implies\frac{x-m}{m}=-\frac{\overline{x}-\overline{m}}{\overline{m}}$ Simplifying we get, $\overline{x}=2\overline{m}-x.\frac{\overline{m}}{m}$ Now $P$ satisfy the condition of $X$. Thus, $\overline{p}=2\overline{m}-p.\frac{\overline{m}}{m}$ Now $A$,$P$,$B$ collinear . So $\frac{p-a}{b-a}=\frac{\overline{p}-\overline{a}}{\overline{b}-\overline{a}}$ $\implies \frac{p-a}{b-a}=\frac{\overline{p}-\frac{1}{a}}{\frac{1}{b}-\frac{1}{a}}$ $\implies p+\overline{p}ab=a+b$ $ \implies p+ab(2\overline{m}-p.\frac{\overline{m}}{m})=a+b$ $\implies p=\frac{m(a+b-2ab\overline{m})}{m-ab\overline{m}}$ Similarly, $\implies q=\frac{m(a+c-2ac\overline{m})}{m-ac\overline{m}}$ Now $\overline{m}=\frac{2}{a}+\frac{1}{b}+\frac{1}{c}=\frac{2bc+ac+ab}{2abc}$ Again, $m-ac\overline{m}=\frac{2a+b+c}{2} - \frac{2bc+ac+ab}{2b}= \frac{(a+b)(b-c)}{2b}$ Now, $a+c-2ac\overline{m}=a+c-\frac{2bc+ac+ab}{b}= \frac{-c(a+b)}{b}$ Thus, $q= \frac{m(\frac{-c(a+b)}{b})}{\frac{(a+b)(b-c)}{2b}} = - \frac{2mc}{b-c}$ Simillarly, $p=- \frac{2mb}{c-b}$ Now, $OP=|o-p|=|p|=|\frac{-2mb}{c-b}|=|\frac{2mb}{c-b}|=|b|.|\frac{2m}{b-c}|=1.|\frac{2m}{b-c}|=|c|.|\frac{2m}{b-c}|=|-\frac{2mc}{b-c}|=|q|=|-q|=|o-q|=OQ$ Thus $OP=OQ$ and we are done .

It's probably butterfly theorem.

Here is a stylish solution , redefine $P,Q$ as points in $AB,AC$ respectively such that $APHQ$ is a paralelogram (by throwing parallels from $H$ we can clearly see those points are unique so we are fine). Hence by the parallels we must have $\angle BHP=90=\angle QHC$ hence in $BPQC$ the point $H$ has an isogonal conjugate, and it is in fact $O$ as in $\triangle ABC$ we have $H,O$ isogonal conjugates, now: $$\angle QPO=\angle BPH=\angle BAC=\angle HQC=\angle PQO \implies \triangle POQ \; \text{isosceles} \implies OM \perp PQ$$Therefore we have that in fact $P,Q$ are the points given in this problem as trivially now $MP=MQ$ we are done .