Q. 53.8( 435 Votes )

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

Answer :

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.

It is given that the diagonals of ABCD are equal and bisect each other at right angles.

Therefore, AC = BD, OA = OC, OB = OD, and

∠AOB = ∠BOC = ∠COD = ∠AOD = 90^{0}

**To prove:** ABCD is a square,**Proof:**

we have to prove that ABCD is a parallelogram with all of its sides equal and one of the interior angle is right angle.

**AB = BC = CD = AD, and one of its interior angles is 90**

**or,**

^{0}

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

∠AOB = ∠ COD (Vertically opposite angles)

**SAS**) Definition: Triangles are

**congruent**if any pair of corresponding sides and their included angles are equal in both triangles.

ΔAOB ΔCOD (SAS congruence rule)

AB = CD (By CPCT) ......... eq(1)

And,

**∠OAB = ∠****OCD** (By CPCT)

However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel

**AB || CD** ............................eq(2)

From equations (1) and (2), we obtain ABCD is a parallelogram

**( Theorem: Quadrilateral with any of opposite sides equal and parallel is a parallelogram.) **

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠ AOD = ∠ COD (Given that each is 90^{0})

OD = OD (Common)

ΔAOD ΔCOD (SAS congruence rule)

AD = DC (3)

However,

AD = BC and

AB = CD (Opposite sides of parallelogram ABCD)

AB = BC = CD = DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ΔADC and ΔBCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

ΔADC ΔBCD (SSS Congruence rule)

∠ADC = ∠BCD (By CPCT)

However,

∠ADC + ∠BCD = 180^{0} (Co-interior angles)

∠ADC + ∠ADC = 180^{0}

2 ∠ADC = 180^{0}

∠ADC = 90^{0} One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90^{0}

Therefore, ABCD is a square

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