Question

please please please help me out to solve this question please​

Answer 1

Step-by-step explanation:

Consider △ADP & △QBC

∠D=∠B (Opposite side of parallelogram.)

∴DA=BC (Opposite side of parallelogram.)

∠x=∠x [bisector of opposite angle.]

∴ By ASA congruence rule. [△ADP≅△QBP]

then by CPCT, DP=QB⟶(1)

∠x=∠x [bisector of opposite angle.]

△ADP≅△QBP [by ASA]

AB=DC [opposite side of a parallelogram.]

AQ+QB=DP+PC [QB=DP from 1]

AQ=PC⟶(2)

Hence distance between AP and QC are equal at all place hence proved AP∥QC⟶(3)

∴AQCP is parallelogram.