If APB and CQD are two parallel lines,then th | vedclass

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(A) Let $PN$ be the bisector of $\angle APQ$ and $QN$ be the bisector of $\angle CQP$.Since $APB \parallel CQD$,the sum of consecutive interior angles

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a rectangle

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(A) Let $PN$ be the bisector of $\angle APQ$ and $QN$ be the bisector of $\angle CQP$.Since $APB \parallel CQD$,the sum of consecutive interior angles is $\angle APQ + \angle CQP = 180^\circ$.Dividing by $2$,we get $\frac{1}{2}\angle APQ + \frac{1}{2}\angle CQP = 90^\circ$.In $	riangle PNQ$,$\angle NPQ + \angle NQP = 90^\circ$,therefore $\angle PNQ = 90^\circ$.Similarly,for the other vertices $M, P, Q$,we can show that all angles of the quadrilateral $PNQM$ are $90^\circ$.Thus,the quadrilateral $PNQM$ formed by the angle bisectors is a rectangle.

If $APB$ and $CQD$ are two parallel lines,then the bisectors of the angles $APQ, BPQ, CQP$ and $PQD$ form

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