Answer :


Given ABCD is a parallelogram in which AB||CD and BC||AD.


A=90°


Since, DC||AB,


D+A=180° (co-interior angles)


D+90°=180°


D=180°-90°


D=90°…………..(1)


Since, AD||BC,


B+A=180° (co-interior angles)


B+90°=180°


B=180°-90°


B=90°…………..(2)


Also, D+C=180° (co-interior angles)


C+90°=180° (from(1))


C=180°-90°


C=90°…………..(3)


Hence, if each angle of a parallelogram is 90° then all the other angles of the parallelogram are 90°.


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