Prove that, the bisector of any two consecutive angles of parallelogram intersect at right angle. -Maths 9th

Prove that, the bisector of any two consecutive angles of parallelogram intersect at right angle. -Maths 9th
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Construct a Parallelogram ABCD where AB is Parallel to CD and BC is Parallel to AD. Make Angle Bisectors of Angle A and Angle B and let them join at O and let Angle OAB be X and Angle OBA be Y.... Angle A+ Angle B =180°(£q.1) (Since sum of adjacent angles of a Parallelogram is 180°). . . . Now multiply £q 1 by 1/2 1/2AngleA+1/2AngleB=90° ANGLE OAB+ ANGLE OBA=90° X+Y=90°...(£q.2) X+Y+Angle AOB =180°...(£q.3)(angle sum property of a triangle) Since X+Y=90°(£q.2) Angle AOB=180°-90° = 90° Hence Proved that  the bisector of any two consecutive angles of parallelogram intersect at right angle I HOPE IT HELPS
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