This problem was proposed by Burii.

Cleary its enough to show that $\odot(BAD)$ is tangent to $AC$. Thus we get, Rephrased Problem wrote: Let $\omega_1,\omega_2$ be two circles intersecting at $\{A,B\}$. Tangents to $\omega_1,\omega_2$ at $A$ intersect $\omega_2,\omega_1$ for the second time at $D$ and $C$ respectively. Let $X$ and $Y$ be projections of $B$ on $AD, AC$. Prove that the orthocenter of $\triangle BXY$ lies on $DC$. Obviously $H$ is the reflection of $A$ about midpoint of $XY$ thus $AXHY$ is a parallelogram. Define $H':=\ell \cap DC$ where $\ell$ is a line parallel to $AC$ passing through $X$. Thus, its enough to show $H'Y\|AD$. For that, we note that $\triangle BAY\sim\triangle BDX$ and $\triangle BXA\sim\triangle BYC$. Thus, $$\frac{DH'}{H'C}=\frac{DX}{XA}=\frac{AY}{YC}\implies H'Y\| AD\implies H'\equiv H\quad\blacksquare$$

Jupiter_is_BIG wrote: Cleary its enough to show that $\odot(BAD)$ is tangent to $AC$. Thus we get, Rephrased Problem wrote: Let $\omega_1,\omega_2$ be two circles intersecting at $\{A,B\}$. Tangents to $\omega_1,\omega_2$ at $A$ intersect $\omega_2,\omega_1$ for the second time at $D$ and $C$ respectively. Let $X$ and $Y$ be projections of $B$ on $AD, AC$. Prove that the orthocenter of $\triangle BXY$ lies on $DC$. Obviously $H$ is the reflection of $A$ about midpoint of $XY$ thus $AXHY$ is a parallelogram. Define $H':=\ell \cap DC$ where $\ell$ is a line parallel to $AC$ passing through $X$. Thus, its enough to show $H'Y\|AD$. For that, we note that $\triangle BAY\sim\triangle BDX$ and $\triangle BXA\sim\triangle BYC$. Thus, $$\frac{DH'}{H'C}=\frac{DX}{XA}=\frac{AY}{YC}\implies H'Y\| AD\implies H'\equiv H\quad\blacksquare$$ Good sol: Great that you used the 'same' method (I totally made that up)