Question

Prove that the bisector of interior angles of a parallelogram form a rectangle.

Answer 1

To prove: MNOP is a rectangle.

In parallelogram $ABCD$

$\angle A=\angle D={90}^{\circ }$ [they form a straight line]

$\therefore IN△AMD,\angle M={90}^{\circ }$

$\angle M=\angle N={90}^{\circ }$ [they form a straight line]

Similarly,

$\angle M=\angle P={90}^{\circ }$

And

$\angle P=\angle O={90}^{\circ }$

$\therefore \angle MPO=\angle PON\angle ONM=\angle NMO={90}^{\circ }$

$\therefore$ MNOP is a rectangle. [A rectangle is a parallelogram with one angle ${90}^{\circ }$]