Since AB is a diameter of the quadrilateral circumcircle (O), O is the midpoint od AB and the angles $\angle ADB = \angle ACB = 90^\circ$ are right. Thus AC, AD are 2 altitudes of the triangle $\triangle ABJ$ intersecting at its orthocenter K and the 3rd altitude $JK \perp AB$. Since the quadrilateral ABCD is cyclic, the intersection of its diagonals K is the pole of the line IJ with respect to the circle (O). Hence $OK \perp IJ$. Since $JK \perp AB \equiv IO$, $OK \perp IJ$, JK and OK are 2 altitudes of the triangle $\triangle IJO$ intersecting at its orthocenter K. The 3rd altitude $IK \perp JO$. Thus we see that if $N \equiv O$ is the midpoint of AB, $IK \perp JN$. Conversely, if N is not the midpoint of AB, then the perpendicular from N to IJ is parallel to OK and it intersects the line JK at a point L different from K. Since $JL \equiv JK \perp AB \equiv IN$ and $NL \perp IJ$, JL and NL are 2 altitudes of the triangle $\triangle IJN$ intersecting at its orthocenter L different from K. Hence, the 3rd altitude $IL \perp JN$ is different from IK. This means that IK is not perpendicular to JN, because IL is.

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Is there a way to prove it without using "pole"?