IMO Hk TST

Let $P$ be the point on the side $AB$ such that $AP = AD$. Then we have
$$BP = AB - AP = AD + BC - AD = BC.$$
Let the internal angle bisector of $\angle ADC$ meet $AB$ at $Q$. We find that
$$\angle QPC = \angle BPC = 90^\circ - \frac{1}{2}\angle CBP = \frac{1}{2}(180^\circ - \angle CBA) = \frac{1}{2}\angle ADC = \angle QDC.$$
This implies $C$, $D$, $P$, $Q$ are concyclic. Similarly, we have
$$\angle DCQ = \angle DPA = 90^\circ - \frac{1}{2}\angle PAD = \frac{1}{2}(180^\circ - \angle BAD) = \frac{1}{2}\angle DCB.$$
This shows $CQ$ bisects $\angle BCD$. Therefore, the internal angle bisectors of $\angle ADC$ and $\angle BCD$ intersect at the point $Q$ on $AB$.