In a parallelogram show that the angle bisectors of 2 adjacent angle intersect at the right angle.
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Let us consider a parallelogram ABCD
Draw the angle bisectors of angle A and angle B
Let their intersection point be O.
Now . We know that /A + /B = 180 degree (Angles on the same side of a transversal are supplementry)
Divide both sides by 2
1/2 of /A + 1/2 of /B = 90 degree
=> /OBA + /OAB = 90 degree --- Eqn 1
In triangle AOB
/OBA + /AOB + /OAB = 180 degree( Angle sum property of a triangle)
/AOB + 90 degree = 180 degree ( From Eqn 1)
/AOB = 180 - 90 degree
=> /AOB = 90 degree
Hence, proved that in a parrallelogram angle bisectors of 2 adjacent angles intersect at a right angle
Draw the angle bisectors of angle A and angle B
Let their intersection point be O.
Now . We know that /A + /B = 180 degree (Angles on the same side of a transversal are supplementry)
Divide both sides by 2
1/2 of /A + 1/2 of /B = 90 degree
=> /OBA + /OAB = 90 degree --- Eqn 1
In triangle AOB
/OBA + /AOB + /OAB = 180 degree( Angle sum property of a triangle)
/AOB + 90 degree = 180 degree ( From Eqn 1)
/AOB = 180 - 90 degree
=> /AOB = 90 degree
Hence, proved that in a parrallelogram angle bisectors of 2 adjacent angles intersect at a right angle
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