how to find this angle inside a quadrilateral

Question

in this quadrilateral, $AB=CD$ and they intersect at $I$. perpendicular bisectors of $AC$ and $DB$ meet at poink $k$, perpendicular bisectors of $AD$ and $BC$ meet at point $J$. what is $\angle KIJ$?

in geogebra the circumcircles of triangles $AID$ and $BIC$ intersect at $J$.but i don't know how to prove this

Answer 1

Observe that , $\triangle KAB\cong \triangle KCD$ and $\triangle JBA\cong \triangle JCD$ by $S-S-S$ criterion of congruence.

Now, in quadrilateral $ACIK$, $\angle KCI=\angle KCD=\angle KAB=\angle KAI$ and thereafter it is cyclic. Similarly, quadrilateral $ICBJ$ is cyclic.

Now, by simple angle chasing, show that $KI$ bisects $\angle AID$ and $JI$ bisects $\angle DIB$ and therefore $\angle KIJ=90^{\circ}$.