Show that the bisectors of angles of a parallelogram form a rectangle.
Given: A parallelogram \(ABCD\).
\(AP\), \(BP\), \(CR\) and \(DR\) are bisectors of \(A\), \(B\), \(C\) and \(D\) respectively.
The intersection of \(DR\) and \(AP\) is \(S\), and \(BP\) and \(CR\) is \(Q\).
Since \(ABCD\) is a parallelogram, \(AB || DC\)
Adjacent angles of a parallelogram are .
In \(ΔAPB\), \(∠DAB + ∠CBA =\) \(^\circ\)
\(\frac{1}{2}∠DAB + \frac{1}{2} ∠CBA = \frac{1}{2}\times\) \(^\circ\)
\(∠PAB + ∠PBA = 90^\circ\) []
Sum of the angles of a triangle is \(180^\circ\).
\(∠APB + ∠PAB + ∠ PBA = 180^\circ\)
\(∠APB + 90^\circ = 180^\circ\)
\(∠APB = 180^\circ - 90^\circ\)
\(∠APB =\) \(^\circ\)
Similarly, we can prove \(∠DRC = 90^\circ\), \(∠PSR = 90^\circ\) and \(∠PQR = 90^\circ\).
So, \(PQRS\) is a quadrilateral with all the internal angles \(90^\circ\).
Thus, \(PQRS\) is a rectangle.
Hence, proved.