Lemma. If $B C D$ is the isosceles triangle which is outside the triangle $A B C$ and has

$$\measuredangle C B D=\measuredangle B C D=90^{\circ}-\alpha: \stackrel{\text { not }}{=} \beta,$$

then $A D \perp B_{A} C_{A}$.
Proof of the lemma. Construct an isosceles triangle $A B E$ outside the triangle $A B C$, so that $\measuredangle A B E=\measuredangle A E B=\beta$.
Then $A E=A B=A B_{A}$ and $\measuredangle E A B_{A}=\alpha$, so a rotation of center $A$ and angle $\alpha$ sends $C_{A}$ to $C$ and $B_{A}$ to $E$, hence $\Varangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{E C}\right)=\alpha$ (the angle between vectors is considered oriented). Also triangles $E B A$ and $B C D$ are similar, so a rotation of center $B$ and angle $\beta$, followed by
!
a dilation of ratio $\frac{E B}{A B}=\frac{B C}{B D}$ sends $E$ to $A$ and $C$ to $D$, hence $\measuredangle(\overrightarrow{E C}, \overrightarrow{A D})=\beta$ (also oriented angle).
This shows that

$$\measuredangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{A D}\right)=\measuredangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{E C}\right)+\measuredangle(\overrightarrow{E C}, \overrightarrow{A D})=\alpha+\beta=90^{\circ} .$$

Returning to the solution of the problem, denote $A^{\prime}$ the intersection of $B C$ with the perpendicular from $A$ to $B_{A} C_{A}$. Then $A^{\prime}$ belongs to the segment $B C$ and

$$\frac{A^{\prime} B}{A^{\prime} C}=\frac{A B \sin (B+\beta)}{A C \sin (C+\beta)}$$

Since similar relations are true for the intersections $B^{\prime}, C^{\prime}$ of the other two perpendiculars with the opposite sides, this yields

$$\frac{A^{\prime} B}{A^{\prime} C} \cdot \frac{B^{\prime} C}{B^{\prime} A} \cdot \frac{C^{\prime} A}{C^{\prime} B}=\frac{A B \sin (B+\beta)}{A C \sin (C+\beta)} \cdot \frac{B C \sin (C+\beta)}{B A \sin (A+\beta)} \cdot \frac{C A \sin (A+\beta)}{C B \sin (B+\beta)}=1$$

whence the conclusion.
Remark. The conditions 'acute-angled' and ' $\alpha<90^{\circ}$ ' are not essential, but without them there are cases when $A^{\prime}$ does not belong to the segment $B C$, or the perpendiculars become parallel.