# Prove that angles of rectangle are right angles

$Let\;ABCD\;be\;a\;rectangle.\\* We\;have\;to\;prove\;that\; \angle A=\angle B=\angle C=\angle D=90^{\circ}\\* We\;know\;that\;a\;rectangle\;is\; parallelogram\;where\;its\;one\;angle\;is\;90^{\circ}\\* Let\;us\;assume\; \angle A=90^{\circ}\\*Now,\\*\Rightarrow AD\parallel BC \;\;\;\;\;\;\;\;\;\;\;\;\;\;[opposite\;sides\;of\;a\;parallelogram\;are\;parallel]\\*And\;AB\;is\; transversal.\\* \because interior\;angles\;on\;the\;same\;side\;of\;a\; transversal\;are\;supplementary\\* \Rightarrow \angle A+\angle B = 180^{\circ}\\* \Rightarrow 90^{\circ}+\angle B=180^{\circ}\;\;\;\;\;\;\;\;\;\;\;(\because we\;assumed\;\angle A=90^{\circ})\\* \Rightarrow \angle B=90^{\circ}\\* Again,\;we\;know\;that \;opposite\;angles\;of\;a\; parallelogram \;are \;equal.\\* \therefore \angle C=\angle A \;and \;\angle D=\angle B\\* \therefore \angle C=90^{\circ}\;and\;\angle D=90^{\circ}\\* \therefore \angle A=\angle B=\angle C=\angle D=90^{\circ}\\* Hence,\;it\;is\;proved\;that\;ABCD\;is\;a\;rectangle.$

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