A Few Rectangles

1.Observation: It is easy to check that $AEBH$ and $AECF$ are cyclic. $J, K, L$ are the mid-points of the diagonals $(AC, BD)$ , $AF, DH$ of $ABCD, AEFD$ and $DEHI$ . $\therefore JK||CF$ and $JL||BH$ .

2.Angle Chasing: (Draw a clear picture to follow the chasing) $\angle DJL=\angle DBH=\angle DBA+\angle ABH=\angle DBA+\angle AEH$ $=\angle DBA+(90^\circ-\angle HAE)=\angle DBA+(90^\circ-\angle AED)$ $[\because IH||DE]$ $=\angle DBA+(90^\circ-25^\circ)=\angle DBA+65^\circ$

Similarly, $\angle AJK=\angle ACD+65^\circ$

Now, $\angle AJK+\angle DJL=$ $(\angle AJD+\angle DJK)+(\angle DJA+\angle AJL)$ $=(\angle AJD+\angle DJK+\angle AJL)+\angle DJA=$ $\angle KJL+\angle DJA.$

$\therefore \angle KJL=(\angle AJK+\angle DJL)-\angle DJA$ $=(130^\circ+\angle DBA+\angle ACD)-\angle DJA=$ $130^\circ+(2\angle DBA-\angle DJA)=\boxed{130^\circ}$ .