Question

ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that: 

(i) ABCD is a square 

(ii) diagonal BD bisects ∠B as well as ∠D

Answer 1

Given: ABCD is a rectangle.

Step 1: Angle Properties of a Rectangle

We are given that ABCD is a rectangle, and it is known that opposite angles in a rectangle are equal. Thus, we have:

\[ \angle A = \angle C \] 
Since angles \( \angle A \) and \( \angle C \) are equal, it follows that:

 

\[ \frac{1}{2} \angle A = \frac{1}{2} \angle C \] Hence:

 

\[ \angle A = \angle C \]

Step 2: Bisecting Angles and Equal Sides

Next, we observe that diagonal \( AC \) bisects both angles \( \angle A \) and \( \angle C \). Therefore, the angle \( \angle DCA \) is equal to \( \angle DCA \) by the reflexive property. This implies:

\[ \angle DCA = \angle DCA \quad (\text{By reflexive property}) \]Additionally, since \( \angle DCA = \angle DCA \), we conclude:

 

\[ CD = DA \quad (\text{Sides opposite to equal angles are also equal}) \] Given that \( DA = BC \) and \( AB = CD \) (since opposite sides of a rectangle are equal), we have:

 

\[ AB = BC = CD = DA \] Thus, ABCD is a rectangle where all the sides are equal, which means ABCD is a square. Hence, we have:

 

\[ ABCD \text{ is a square}. \]

Step 3: Diagonal Bisects Angles

Now, let us join diagonal \( BD \). In triangle \( BCD \), we know that:

\[ BC = CD \quad (\text{Sides of a square are equal to each other}) \] \p>Since \( BC = CD \), we can conclude that:

 

\[ \angle CDB = \angle CBD \quad (\text{Angles opposite to equal sides are equal}) \] However, it is also given that \( \angle CDB = \angle BD \) (Alternate interior angles for \( AB \parallel CD \)), and \( \angle CBD = \angle ABD \). Thus:

 

\[ \angle CDB = \angle ABD \] 
Since the diagonal \( BD \) bisects angle \( \angle B \), we also conclude that:

 

\[ \angle CBD = \angle ABD \quad (\text{By the angle bisector property}) \] 

Moreover, since \( \angle CBD = \angle ADB \) (Alternate interior angles for \( BC \parallel AD \)), we finally conclude that:

 

\[ \angle CDB = \angle ABD \]

Conclusion:

From the above steps, we have shown that diagonal \( BD \) bisects the angles, and all sides of the rectangle are equal. Hence, ABCD is a square.