Hkg TST

This is a special case of Jacobi's theorem. Alternatively, by Ceva's theorem, it suffices to prove
$$\frac{\sin \angle ACX}{\sin \angle BCX} \times \frac{\sin \angle BAY}{\sin \angle CAY} \times \frac{\sin \angle CBZ}{\sin \angle ABZ} = 1.$$
Applying the sine law to $\triangle ACX$ and $\triangle BCX$, we obtain
$$\sin \angle ACX = AX \times \frac{\sin \angle CAX}{CX}, \quad \sin \angle BCX = BX \times \frac{\sin \angle CBX}{CX}.$$

Combining these, since $AX = BX$, we get
$$\frac{\sin \angle ACX}{\sin \angle BCX} = \frac{\sin \angle CAX}{\sin \angle CBX}.$$
Similarly, we find that
$$\frac{\sin \angle ACX}{\sin \angle BCX} \times \frac{\sin \angle BAY}{\sin \angle CAY} \times \frac{\sin \angle CBZ}{\sin \angle ABZ} = \frac{\sin \angle CAX}{\sin \angle CBX} \times \frac{\sin \angle ABY}{\sin \angle ACY} \times \frac{\sin \angle BCZ}{\sin \angle BAZ}.$$
Note that $\angle CAX = \angle BAZ$, $\angle ABY = \angle CBX$ and $\angle BCZ = \angle ACY$. Therefore, this is equal to 1, and we are done.