**Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.**

**Solution:**

Let's suppose that we have a quadrilateral ABCD with AO=OC, BO=OD, AC=BD and its diagonals AC and BD meet each other at .

We need to show that it is a square.

In and

AO=CO ** (Given)**

**(Each given equal to )**

BO=DO **(Given)**

Therefore, **by SAS congruence rule**, .

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

But, these are alternate interior angles which means that . **(1)**

Similarly, we can prove that ** (2)**

From **(1)** and **(2)**, we can say that quadrilateral ABCD is a parallelogram. Hence, we have AB=CD and BC=AD because opposite sides of a parallelogram are equal. ** (3)**

Now, in and

AO=AO **(Common)**

** (Each given equal to )**

OB=OD **(Given)**

Therefore, **by SAS congruence rule**, we have

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

In and

AC=BD **(Given)**

AD=BC **(Proved above in (1) )**

CD=DC ** (Common)**

Therefore, **by SSS congruence rule**,

** (Corresponding parts of congruent triangles are equal) (5)**

But, we also have ** (Co-Interior angles) (6)**

From **(5)** and **(6)**, we can say that

** (7)**

From **(3)**, **(4)** and **(7)**, we can say that ABCD is a parallelogram having all the sides equal and we have showed that it's one angle is equal to which is enough to consider it a square. Therefore, ABCD is a square.