The Bisectors of any Two Consecutive Angles of a Parallelogram Intersect at Right Angle


 
 
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The Bisectors of any Two Consecutive Angles of a Parallelogram Intersect at Right Angle

The Bisectors of Any Two Consecutive Angles Intersect at Right Angle: We know that parallelogram is a quadrilateral in which both the opposite pair of sides are parallel and equal to each other, The bisectors of any two consecutive angles intersect at right angle.

Theorem 5: In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.

GIVEN A parallelogram ABCD such that the bisectors of consecutive angles A and B intersect at P.

To prove   <APB = 90

Proof  Since ABCD is a parallelogram. Therefore,   AD large parallel BC

Now, AD large parallel BC and transversal AB intersects them.

large therefore      <A + <B = 180           [ large because Sum of consecutive interior angles is 180]

large Rightarrow;frac{1}{2}angle A+frac{1}{2}angle B = 90^0

large Rightarrow angle BAO + angle ABO = 90^0...(i)   [large because AO and BO are bisectors of <A and <B]

In large DeltaAPB, we have

     <1 + <APB + <2 = 180

large Rightarrow  90 + <APB = 180                      [ From (i), <1 + <2 = 90]

large Rightarrow      <APB = 90

ILLUSTRATION:  In the given figure, AP and BP are the bisectors of angleA and angleB which meet at P inside the parallelogram ABCD. Prove that 2angleAPB = angleC + angleD.

Solution: In a parallelogram the angle bisectors of a parallelogram meet at right angles.

therefore ;;angle APB = 90^0      ...........(1)

As ABCD is a parallelogram. Therefore,   AD large parallel BC

Now, AD large parallel BC and transversal CD intersects them.

large therefore      angle C + angle D = 180^0           [ large because Sum of consecutive interior angles is 180]

Rightarrow angle C + angle D = 2 times 90^0

Rightarrow angle C + angle D = 2 times angle APB      [From Eq. 1]

Hence Proved

 

 
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