Question

Show that the diagonals of a square are equal and bisect each other at right angles.​

Answer 1

$\huge\sf\underline\red{Given:-}$

The diagonals of a square are equal and bisect each other at right angles.

$\huge\sf\underline\pink{Explanation:-}$

Let ABCD be a square and its diagonals AC and BD intersect each other at O.

$\huge\sf\underline\purple{To\:show\:that:-}$

AC = BD

AO = OC

and ∠AOB = 90°

_________________

$\huge\sf\underline\orange{Proof:-}$

In ΔABC and ΔBAD,

BC = BA (Common)

∠ABC = ∠BAD = 90°

AC = AD (Given)

, ΔABC ≅ ΔBAD [SAS congruency]

_________________

Thus,

AC = BD [CPCT]

Diagonals are equal.

Now,

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

ΔAOB ≅ ΔCOD [AAS congruency]

_________________

Thus,

AO = CO [CPCT].

Diagonal bisect each other.

Now,

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

ΔAOB ≅ ΔCOB [SSS congruency]

_________________

also, ∠AOB = ∠COB

∠AOB + ∠COB = 180° (Linear pair)

Thus, ∠AOB = ∠COB = 90°

$\small\bold\red{Diagonals\:bisect\:each\:other\:at\: right\:angles}$

Answer 2

$\huge{\orange{\underline{\purple{\mathscr{Solution}}}}}$

In ΔABC and ΔBAD,

BC = BA (Common)

∠ABC = ∠BAD = 90°

AC = AD (Given)

, ΔABC ≅ ΔBAD [SAS congruency]

_________________

Thus,

AC = BD [CPCT]

Diagonals are equal.

Now,

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

ΔAOB ≅ ΔCOD [AAS congruency]

_________________

Thus,

AO = CO [CPCT].

Diagonal bisect each other.

Now,

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

ΔAOB ≅ ΔCOB [SSS congruency]

_________________

also, ∠AOB = ∠COB

∠AOB + ∠COB = 180° (Linear pair)

Thus, ∠AOB = ∠COB = 90°