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In a parallelogram ABCD the bisector os the consecutive angles A and B intesect at p show that angle APB=90�
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In a parallelogram ABCD the bisector os the consecutive angles A and B...
Given in parallelogram ABCD AO and BO are bisectors of angle A and angle B respectively.

We know that sum of adjacent angles in a parallelogram is 180degree

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In a parallelogram ABCD the bisector os the consecutive angles A and B...
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In a parallelogram ABCD the bisector os the consecutive angles A and B...

Proof that angle APB is 90° in a parallelogram ABCD:

1. Given Information:
- ABCD is a parallelogram.
- Angle APC bisects angle A.
- Angle BPD bisects angle B.
- The bisectors of consecutive angles A and B intersect at point P.

2. Property of Bisectors:
- The angle bisectors of a parallelogram divide the opposite sides into equal segments.

3. Proof:
- In parallelogram ABCD, angle APC and angle BPD divide sides AD and BC equally.
- Let AD = a, AP = x, PD = x, BC = b, PB = y, and PC = y.
- By angle bisector theorem, we have:
x/y = a/b (from triangle APD and triangle CPB)
- Since ABCD is a parallelogram, a = b.
- Therefore, x = y.

4. Conclusion:
- In triangle APB, we have AP = PB.
- This implies that triangle APB is an isosceles triangle with AP = PB.
- In an isosceles triangle, the angles opposite the equal sides are equal.
- Therefore, angle APB = angle PAB + angle PBA = 90° (since sum of angles in a triangle is 180°).

5. Therefore, angle APB is 90° in parallelogram ABCD.
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In a parallelogram ABCD the bisector os the consecutive angles A and B intesect at p show that angle APB=90�?
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