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Now extend PR to TU with T on (A) and U on (B). Let TM and UN meet in V.
Note that angle(MAN) = angle(MBN) and thus angle(VTU) = angle(MTP) = angle(MAN)/2 = angle(MBN)/2 = angle(RUN) = angle(VUT), so triangle TUV is isosceles with VT=VU.
We also see that angle(MBN) = angle(VTU) + angle(TUV) so that angle(MBN) and angle(NVM) add to 180 degrees, so that V lies on the circle with diameter AB.
Now note that angle(TVA)=angle(KVA), as these angles intercept congruent chords, and angle(UVB) = angle(LVB). By reflection through AV and BV respectively, we see that in fact VT=VK=VL=VU.
Let W be the reflection of V through AB. Triangle WMN is isosceles just as triangle KLV. Hence angle(WVN) = angle(WMN) = angle(WNM) = angle(WVM), so VW bisects angle(TVU). So VW is perpendicular to PR as angle bisector of the apex of an isosceles triangle. And thus PR is parallel to AB. Similarly QS is parallel to AB. And PQSR is a rectangle.
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