China Mathematical Competition (Complementary Test)

By reduction to absurdity, assume that $A, B, D, C$ are not concyclic. Let the circumcircle (with radius $r$) of $ABC$ intersect $AD$ at point $E$. Join $BE$ and extend it to intersect line $AN$ at point $Q$; join $CE$ and extend it to intersect $AM$ at $P$. Join $PQ$, as seen in Fig. 1.2.

Fig. 1.2

We have
$$\begin{aligned} PK^2 &= \text{the power of } P \text{ with respect to } \odot O + \\ &\quad \text{the power of } Q \text{ with respect to } \odot O \\ &= (PO^2 - r^2) + (KO^2 - r^2). \end{aligned}$$
(We will prove it in the appendix)

In the same way,
$$QK^2 = (QO^2 - r^2) + (KO^2 - r^2).$$
Then we have
$$PO^2 - PK^2 = QO^2 - QK^2.$$
Therefore, $OK \perp PQ$. By the given condition $OK \perp MN$, we get that $PQ \parallel MN$. Then we have
$$\frac{AQ}{QN} = \frac{AP}{PM}. \qquad \textcircled{1}$$
By Menelaus' Theorem, we obtain
$$\frac{NB}{BD} \cdot \frac{DE}{EA} \cdot \frac{AQ}{QN} = 1, \qquad \textcircled{2}$$
$$\frac{MC}{CD} \cdot \frac{DE}{EA} \cdot \frac{AP}{PM} = 1. \qquad \textcircled{3}$$
From ①, ②, ③, we get $\frac{NB}{BD} = \frac{MC}{CD}$, or $\frac{ND}{BD} = \frac{MD}{DC}$. Then $\triangle DMN \sim \triangle DCB$, which implies $\angle DMN = \angle DCB$. Then $BC \parallel MN$. Therefore, $OK \perp BC$, and that means $K$ is the midpoint of $BC$, which is a contradiction. This completes the proof that $A, B, D, C$ are concyclic.