**ABCD is a rectangle in which diagonal AC bisects as well as . Show that:**

**(i)** ABCD is a square **(ii)** diagonal BD bisects as well as .

**Solution (i)**

Let's suppose that we have rectangle ABCD in which diagonal AC bisects as well as .

We have AB=CD and BC=AD **(Opposite sides of rectangle are equal) (1)**

In and

** (Given)**

AC=AC **(Common)**

**(Given)**

Therefore, **by ASA congruence rule**,

** (Corresponding parts of congruent triangles are equal) (2)**

From **(1)** and **(2)**, we can say that ABCD is a rectangle having all the sides equal. It means that ABCD is a square.

**Solution (ii)**

In solution **(i)**, we have showed that ABCD is a square.

Now in and

BC=BA **(Sides of square are equal)**

BD=BD **(Common)**

CD=AD ** (Sides of square are equal)**

Therefore, **by SSS congruence rule**,

**(Corresponding parts of congruent triangles are equal) (3)**

And, **(Corresponding parts of congruent triangles are equal) (4)**

From **(3)** and **(4)**, we can say that BD bisects as well as .