Question

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

Answer 1

Solution :

Let ABCD be a square such that that its diagonals AC and BD intersect at O. We have to prove that AC = BD and AC and BD bisect each other at right angle.

In ∆'s BAD and CDA,we have :

• BA = CD

• $\sf\angle\: BAD = \angle\: CDA$ [Each angle equals 90°]

• AD = DA[(common]

By SAS congruence criterion,we obtain :

$\sf \: \triangle \: BAD \cong \: CDA$

$\longrightarrow \sf \: BD = CA$

We know that :

→ Every square is a parallelogram and diagonals of a parallelogram bisect each other.

$\therefore \sf \: OA = OC \: and \: OB =O D$

If we conside triangles AOB and COB. We'll have :

• AB = CB [ABCD is a square]

• OB = OB [Common]

• OA = OC

By SSS congruence criterion,we obtain :

$\sf \: \triangle \: AOB \cong COB$

$\longrightarrow \sf \: \angle \: AOB = \angle \: COB$

But,

$\longrightarrow \sf \: \angle \: AOB + \angle \: COB = {180}^{ \circ}$

$\therefore \sf \: \angle \: AOB = \angle \: COB = {90}^{ \circ}$

Hence, Diagonals of square ABCD are equal and bisect each other at right angles.

Answer 2

Given :-

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

To prove :-

The diagonals of the square ABCD are equal and bisect each other at right angles. i.e.,

AC = BD, AO = OC, BO = BD, and ∠AOB = 90°.

Proof :-

In $\Delta$ ABC and $\Delta$ BDA,

AD = BC             (as ABCD is a square)

∠ABC=∠BAD    ( = 90°)

AB = BA             (side in common)

∴ By SSS congruence criteria, $\Delta$ ABC ≅ $\Delta$ BDA.

So, BC = AC    (by CPCT)

Thus, it is proved that the diagonals of a square are equal.

In $\Delta$ OAD and $\Delta$ OBC,

AD = BC             (as ABCD is a square)

∠OAD=∠OCB    (transversal AC between DA || CB)

∠ODA=∠OBC    (transversal BD between BC || DA)

∴ By ASA congruence criteria, $\Delta$ ABC ≅ $\Delta$ BDA.

OA = OC             (by CPCT)    ...(i)

OB = OD             (by CPCT)    ...(ii)

From (i) and (ii), it is proved that that AC and BD bisect each other.

In $\Delta$ OAD and $\Delta$ OAB,

OB = OD             [from (ii)]

AB = AD             (as ABCD is a square)

OA = AO             (side in common)

∴ By SSS congruence criteria, $\Delta$ OAD ≅ $\Delta$ OAB.

$\Delta$ AOB = $\Delta$ AOD            ...(iii)

∠AOB + ∠AOD = 180°  (linear pair)

2 ∠AOB = 180°              [from (iii)]

∠AOB = 90°

∴ Thus it is proved that, the diagonals of a square bisect each other at right angles (90°).