Question

Show that of the diagonals of a quadrilateral
are equal and bisect each other at right-angle
then it is a
square​

Answer 1

Question

Show that of the diagonals of a quadrilteral are equal and bisect each other at right-angle then it is a square.

Given

• Diagonals are equal
• AC = BD. .... (1)
• and The diagonals bisect each other at right angles
• OA = OC; OB = OD. .... (2)
• ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°..... (3)

Proof

Consider triangle AOB and triangle COB

OA = OC. ... (from (2) )

∠AOB = ∠VOH

OB is the common side

Therefore,

Triangle AOB ≅ Triangle COB

From SAS criteria, AB = CB

Similarly we prove

Triangle AOB ≅ triangle DOA, so AB = AD

Triangle BOC ≅ triangle COD, so CB = DC

so, AB = AD = CB = DC ..... (4)

So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is paralleogram.

In Triangle ABC and Triangle DCB

AC = BD

AB = DC

BC is the common side

Triangle ABC ≅ Triangle DCB

$\sf \: so, \: from \: SSS \: criteria,∠ AND = ∠ DCB$

Now,

AB || CD , BC is the transversal

∠B + ∠C = 180°

Answer 2

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.