Theory:

\(PQRS\) is a rectangle in which diagonal \(PR\) bisects \(\angle P\) as well as \(\angle R\). Show that:
 
(i) \(PQRS\) is a square
 
(ii) diagonal \(QS\) bisects \(\angle Q\) as well as \(\angle S\).
 
Solution:
 
A10.png
 
Given: \(PQRS\) is a rectangle and the line segment \(PR\) bisects \(\angle P\) and \(\angle R\).
 
\(\Rightarrow \angle P = \angle R\)
 
\(\Rightarrow \frac{1}{2}\angle P = \frac{1}{2}\angle R\)
 
\(\Rightarrow \angle SPR = \angle SRP\)
 
In \(\Delta SPR\), \(\angle SPR = \angle SRP\)
 
Sides opposite to equal angles are equal.
 
\(PS = SR\) - - - - (I)
 
But, we know that "opposite sides are equal in rectangle".
 
\(PQ = SR\) and \(PS = QR\) - - - - (II)
 
From equation (I) and (II), we get:
 
\(PQ = QR = RS = PS\)
 
\(PQRS\) is a rectangle with all the sides are equal.
 
Thus, \(PQRS\) is a square.
 
 
(ii) Construction: Join the diagonal \(QS\).
 
A11.png
 
In \(\Delta QRS\), \(QR = RS\) (proved above)
 
Angles opposite to equal sides are equal.
 
\(\angle SQR = \angle RSQ\) - - - - (III)
 
Since \(PQ \ || \ RS\) with transversal \(QS\)
 
\(\angle PQS = \angle RSQ\) [alternate interior angles] - - - - (IV)
 
From equation (III) and (IV), we get:
 
\(\angle SQR = \angle PQS\)
 
\(\Rightarrow QS\) bisects \(Q\)
 
Since \(PS \ || \ QR\) with transversal \(QS\).
 
\(\angle PSQ = \angle SQR\) [alternate interior angles] - - - - (V)
 
From equation (III) and (V), we get:
 
\(\angle PSQ = \angle RSQ\)
 
\(\Rightarrow QS\) bisects \(S\)
 
Hence, \(QS\) bisects \(Q\) as well as \(S\).