Question

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

Answer 1

$\huge\mathbb{SOLUTION:-}$

Given:-

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

Explanation:-

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

To show that,

AC = BD

AO = OC

and ∠AOB = 90°

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°

• Diagonals bisect each other at right angles

Answer 2

Step-by-step explanation:

Given:-

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

Explanation:-

(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

AB = BC = CD = AD

Thus, ABCD is a