Let ${\omega }_1$ and ${\omega }_2$ be circles with centres $O_1$ and $O_2$, respectively, such that the radius of ${\omega }_1$ is less than the radius of ${\omega }_2$. Suppose ${\omega }_1$ and ${\omega }_2$ intersect at two distinct points $A$ and $B$. Line $O_1{O}_2$ intersects ${\omega }_1$ at $C$ and ${\omega }_2$ at $D$, such that $C,O_1,O_2,D$ lie on line in that order. Let $K$ be the circumcentre of triangle $O_{1}AD$. Line $KA$ intersects ${\omega }_1$ again at $E$. Line $KB$ intersects ${\omega }_2$ again at $F$.

Let $H$ be the orthocentre of triangle $KAB$. Prove that the line through $H$ parallel to $EF$ is tangent to the circumcircle of triangle $KAB$.

Step 1: $K$ is the circumcenter of $△O_{1}AD$, so $K{O}_1=KA=KD$.

Step 2: $E$ is the second intersection of $KA$ with ${\omega }_1$. Since $A\in {\omega }_1$ and $K$ determines a line through $A$, $E$ is a specific point.

Step 3: $F$ is the second intersection of $KB$ with ${\omega }_2$.

Step 4: $H$ is the orthocenter of $△KAB$. A line through $H$ parallel to $EF$ is tangent to the circumcircle of $△KAB$ iff the distance from the circumcenter of $KAB$ to this line equals the circumradius. This can be verified via angle chasing using the properties of $K$ as circumcenter of $O_{1}AD$.

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