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Question

Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.(shown in the given figure)

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Solution

ABCD is a parallelogram.
BAD+CDA=180° interior angles on the same side of the transversal are supplementary12BAD+12CDA=12×180°SAD+SDA=90° .....1
SAD+SDA+ASD=180° Angle sum property in ASD90°+ASD=180°ASD=90°PSR=90° Vertically opposite angles
Similarly, PQR=90°
DAB+CBA=180° interior angles on the same side of the transversal are supplementary12DAB+12CBA=12×180°BAR+RBA=90° In ABR, BAR+RBA+ARB=180°90°+ARB=180°ARB=90°Similarly,DPC=90°
Thus, PQRS is a rectangle.

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