Question

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

Answer 1

Step-by-step explanation:

Explanation:

______________________________

Given that,

Let ABCD be a quadrilateral

It's iagonals AC and BD bisect each other at right angle at O.

To prove that

The Quadrilateral ABCD is a square.

Proof,

In ΔAOB and ΔCOD,

⇝ AO = CO (Diagonals bisect each other)

⇝ ∠AOB = ∠COD (Vertically opposite)

⇝ OB = OD (Diagonals bisect each other)

⇝ ΔAOB ≅ ΔCOD [SAS congruency]

Thus,

⇝ AB = CD [CPCT] — (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

Now,

⇝ In ΔAOD and ΔCOD,

⇝ AO = CO (Diagonals bisect each other)

⇝ ∠AOD = ∠COD (Vertically opposite)

⇝ OD = OD (Common)

⇝ ΔAOD ≅ ΔCOD [SAS congruency]

Thus,

AD = CD [CPCT] ____ (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB ____ (ii)

also, ∠ADC = ∠BCD [CPCT]

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90° ____ (iii)

One of the interior angles is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

HenceProved!

Answer 2

Step-by-step explanation:

Given that ABCD is a square.

To prove : AC=BD and AC and BD bisect each other at right angles.

Proof:

(i) In a ΔABC and ΔBAD,

AB=AB ( common line)

BC=AD ( opppsite sides of a square)

∠ABC=∠BAD ( = 90° )

ΔABC≅ΔBAD( By SAS property)

AC=BD ( by CPCT).

(ii) In a ΔOAD and ΔOCB,

AD=CB ( opposite sides of a square)

∠OAD=∠OCB ( transversal AC )

∠ODA=∠OBC ( transversal BD )

ΔOAD≅ΔOCB (ASA property)

OA=OC ---------(i)

Similarly OB=OD ----------(ii)

From (i) and (ii) AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB=OD ( from (ii) )

BA=DA

OA=OA ( common line )

ΔAOB=ΔAOD----(iii) ( by CPCT

∠AOB+∠AOD=180° (linear pair)

2∠AOB=180°

∠AOB=∠AOD=90°

∴AC and BD bisect each other at right angles.