show that the angles contained
by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles
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yoir answer is
(i) AD = BC (opposite sides of a parallelogram are equal)
(ii) \(\angle\)DCB = \(\angle\)BAD (opposite angles are equal)
(iii) OC = OA (diagonals of a parallelogram bisect each other)
(iv) \(\angle\)DAB +\(\angle\)CDA = \(180^{\circ}\) (the sum of two adjacent angles of a parallelogram is \(180^{\circ}\) )
yoir answer is
(i) AD = BC (opposite sides of a parallelogram are equal)
(ii) \(\angle\)DCB = \(\angle\)BAD (opposite angles are equal)
(iii) OC = OA (diagonals of a parallelogram bisect each other)
(iv) \(\angle\)DAB +\(\angle\)CDA = \(180^{\circ}\) (the sum of two adjacent angles of a parallelogram is \(180^{\circ}\) )
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