### 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: 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 $$S$$.