For (a), note that the triangles $\triangle{BCB'}$ and $\triangle{MHC}$ are both isosceles. Because furtermore $\sphericalangle{ACH}=\sphericalangle{CBA}$, they are similar; implying that also $\sphericalangle{CMH}=\sphericalangle{CB'H}$ and hence $CMB'H$ is cyclic. Thereto, $\sphericalangle{B'MC}=\sphericalangle{BHC}=\frac{\pi}{2}$ and because $\sphericalangle{ACB}=\frac{\pi}{2}$ too, $BC\| B'M$. For (b), because $\overline{HA}\cdot\overline{HB'}=\overline{HA}\cdot\overline{HB}=\overline{HC}^2$, $A$ is the inverse of $B'$ with respect to the circle centered at $H$ with radius $HC$; therefore $AK$ is tangent to this circle.

Sharygin CR Round Grade 9 2010 P12 wrote: Let $AC$ be the greatest leg of a right triangle $ABC,$ and $CH$ be the altitude to its hypotenuse. The circle of radius $CH$ centered at $H$ intersects $AC$ in point $M.$ Let a point $B'$ be the reflection of $B$ with respect to the point $H.$ The perpendicular to $AB$ erected at $B'$ meets the circle in a point $K$. Prove that a) $B'M \parallel BC$ b) $AK$ is tangent to the circle. Solution: We'll change the question a bit: let the perpendicular to $AB$ at $B'$ intersect the circle at $K_1,K_2$. Now, $$\angle HMC=\angle ACH=\angle CBA=\angle CB'H \implies CHB'M \text{ cyclic} \implies B'M \perp AC \implies B'M||BC$$Let $AD,AE$ be the tangents from $A$ to $\odot (H)$, such that $DC < EC$ $$AD^2=AE^2=AM \cdot AC=AB' \cdot AH \implies AD^2=AB' \cdot AH, ~ \angle ADH=90^{\circ} \implies \angle DB'H=90^{\circ} \text{ and } \angle EB'H=90^{\circ}$$$D - B' - E $ and $DE \perp AB$ at $B'$ $\implies$ $D \equiv K_1$ and $E \equiv K_2$