China Mathematical Olympiad

(1) Since $C$, $A$, $B$, $E$ are concyclic, and $D$, $A$, $B$, $F$ are concyclic, $CA=AD$, and by the theorem of power of a point, we have $$CB\cdot CF=CA\cdot CD=DA\cdot DC=DB\cdot DE.\stackrel{◯}{1}$$ Suppose on the contrary that $l_1$ and $l_2$ do not intersect, then $CD\parallel EF$, hence $\frac{CF}{CB}=\frac{DE}{DB}$. Plugging into (1), we get $C{B}^2=D{E}^2$, thus $CB=DB$, hence $BA\perp CD$. It follows that $CB$ and $DB$ are the diameters of $K_1$ and $K_2$ respectively, hence $K_1$ and $K_2$ have same radii, which contradicts with assumption. Thus $l_1$ and $l_2$ have a unique common point.

(2) Join $AE$, $AF$ and $PF$, we have $$\angle CAE=\angle CBE=\angle DBF=\angle DAF.$$ Since $AP\perp CD$, $AP$ is the bisector of $\angle EAF$. Since $P$ is on the perpendicular bisector of the segment $EF$, $P$ is on the circumcircle of $△AEF$. We have $$\angle EPF=180^{\circ }-\angle EAF=\angle CAE+\angle DAF=2\angle CAE=2\angle CBE.$$ Hence $B$ is on the circle with center $P$ and radius $PE$, denoting this circle by $\Gamma$. Let $R$ be the radius of $\Gamma$. By the

theorem of power of a point, we have $$2C{A}^2=CA\cdot CD=CB\cdot CF=C{P}^2-R^2,$$ thus $$A{P}^2=C{P}^2-C{A}^2=(2C{A}^2+R^2)-C{A}^2=C{A}^2+P{E}^2.$$ It follows that $CA$, $AP$, $PE$ form the side lengths of a right triangle.