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\(PQRS\) 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:
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 \(QS\).
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 \(S\).
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