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  • Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

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Solution.
Given: Let ABCD be a parallelogram and AP, BR, CQ, DS are the bisectors of
\angle A, \angle B, \angle Cand \angle D respectively.
To Prove: Quadrilateral PQRS is a rectangle.

Proof: Since ABCD is a parallelogram
Then DC\parallel AB and DA is transversal.
\angle A+\angle B=180^{\circ}{sum of co-interior angles of a parallelogram}
\frac{1}{2}\angle A+\frac{1}{2}\angle B=90^{\circ} {Dividing both sides by 2}
\angle PAD+\angle PDA=90^{\circ}
\Rightarrow \angle APD=90^{\circ} {\because sum of all the angles of a triangle is 1800}
 \angle QRS=90^{\circ}
Similarly,
\angle RBC+\angle RCB=90^{\circ}
\Rightarrow \angle BRC=90^{\circ}  
\angle QRS=90^{\circ}
Similarly,
\angle QAB+\angle QBA=90^{\circ}
\Rightarrow \angle AQB=90^{\circ}  {\because sum of all the angles of a triangle is 1800}
\angle RQP=90^{\circ} {\because vertically opposite angles}
Similarly,|

\angle SDC+\angle SCD=90^{\circ}
\Rightarrow \angle DSC=90^{\circ} {\because sum of all the angles of a triangle is 1800}
 \angle RSP=90^{\circ} {\because vertically opposite angles}
Thus PQRS is a quadrilateral whose all angles are 90^{\circ}
Hence PQRS is a rectangle.
Hence proved

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