Answer:
Let ABCD be a parallelogram such that adjacent angles \[\angle A\text{ }=\angle B\] Since, \[\angle A\text{ }+\angle B\text{ }=\text{ }180{}^\circ \] \[\Rightarrow \,\,\,\,\,\,\,\,2\angle A\text{ }=\text{ }180{}^\circ \] \[\therefore \,\,\,\,\,\,\,\angle A\text{ }=\angle B\text{ }=\text{ }\frac{180{}^\circ }{2}=90{}^\circ \] Since, opposite angles of a parallelogram are equal. \[\therefore \,\,\,\,\,\angle A\text{ }=\angle C\text{ }=\text{ }90{}^\circ \] and \[\angle B\text{ }=\angle D\text{ }=\text{ }90{}^\circ \] Thus, \[\angle A\text{ }=\text{ }90{}^\circ ,\angle B\text{ }=\text{ }90{}^\circ ,\] \[\angle C\text{ }=\text{ }90{}^\circ and\angle D\text{ }=\text{ }90{}^\circ \]
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