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 $$\begin{array}{rl}P{K}^2& =\text{the power of }P\text{ with respect to }\odot O+\\ & \text{the power of }Q\text{ with respect to }\odot O\\ & =(P{O}^2-r^2)+(K{O}^2-r^2).\end{array}$$ (We will prove it in the appendix)

In the same way, $$Q{K}^2=(Q{O}^2-r^2)+(K{O}^2-r^2).$$ Then we have $$P{O}^2-P{K}^2=Q{O}^2-Q{K}^2.$$ Therefore, $OK\perp PQ$. By the given condition $OK\perp MN$, we get that $PQ\parallel MN$. Then we have $$\frac{AQ}{QN}=\frac{AP}{PM}.\stackrel{◯}{1}$$ By Menelaus' Theorem, we obtain $$\frac{NB}{BD}\cdot \frac{DE}{EA}\cdot \frac{AQ}{QN}=1,\stackrel{◯}{2}$$ $$\frac{MC}{CD}\cdot \frac{DE}{EA}\cdot \frac{AP}{PM}=1.\stackrel{◯}{3}$$ From ①, ②, ③, we get $\frac{NB}{BD}=\frac{MC}{CD}$, or $\frac{ND}{BD}=\frac{MD}{DC}$. Then $△DMN\sim △DCB$, 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.