I will only do the if part since the only if part is very similar. Also I assume you meant the perpendicular bisector of $BC$ with "axis of $BC$". Let $BQ\perp CQ$. We have $BQHC$ is cyclic with center $M$ and so an easy angle chase shows that $\angle BAM = 30$ and in particular, $\triangle BQM$ is equilateral. Let $CQ\cap AB = D$. Since $30 = \angle DAM = \angle DCM$, we have $ADMC$ is cyclic. Now let $(ADH)\cap (HMC) = X'$. By Miquel Point properties we have that $X'MBD$ is also cyclic. We have $\angle DX'B = \angle DMB = \angle BAC = 180 - \angle DX'H \implies X'\in BH$. And since $\angle XHC = 90 = \angle XMC$, we have that $X$ lies on the perpendicular bisector of $BC$ and so $X' = X$ and we are done.

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1' Let $BQ\perp CQ$. We can easily know $XM\perp BC$ as $X$ be the intersection point between $BH$ and the axis of $BC$. So $BQHC$ is cyclic with center $M$. So $\triangle BQM$ is regular triangle as $Q$ be the circumcenter of $(ABM)$. Let$(ACM)\cap (AXH) = P$, we only should prove that $CPQ$ are collinear. That means $\angle QCM=30^{\circ}=\frac{1}{2}\angle BQM=\angle BAM=\angle PCM\Rightarrow CPQ$ are collinear. Then $(ACM)\cap (AXH)\cap CQ = P$. 2' Let $(ACM)\cap (AXH)\cap CQ = P$. We can get a $Q'$ on line $CP$, so that $BQ'\perp CQ'$. Then we can know $BQ’HC$ is cyclic with center $M$. By $\angle QBM=90^{\circ}-\frac{1}{2}\angle BQM= 90^{\circ}-\angle PAM=90^{\circ}-\angle PCM=\angle Q'BM$, we know $Q'=Q$. Then $BQ\perp CQ$. By 1' 2',we have proves this problem.