Question

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

Answer 1

ABCD is a square

R.T.P AC=BC

Proof :- In Triangle ABD, Triangle CBA

AB=BC

AD=BC

angle A=angle B= 90°

By SAS rule

TrianguleABC =~ Triangular CBA

By CPCT AC=BC

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Answer 2

Given: A square.  

To find: The diagonals of a square are equal and bisect each other at right angles  

Solution:

Let ABCD be a square of side x units. All the angles in a square are equal to 90° each. The lines AC and BD form the diagonals of the square. The points A, B and C form a right-angled triangle where AB is the perpendicular, AC is the hypotenuse and BC is the base. Now, the length of AC can be calculated using the Pythagoras theorem.

$AC^{2} = x^{2} + x^{2}$

        $= 2x^{2}$

$AC = x\sqrt{2}$

Similarly, BCD forms a right-angled triangle where BC is the perpendicular, BD is the hypotenuse and CD is the base.

$BD^{2} = x^{2} + x^{2}$

        $= 2x^{2}$

$BD = x\sqrt{2}$

As evident from the calculations, AC and BD, the diagonals, are equal in length.

The diagonals of a square bisect the angles of the square. So, the diagonal BD bisects ∠B and ∠D. Similarly, the diagonal AC bisects ∠A and ∠C. Let the point of intersection of the diagonals be O. Hence,

$\angle ABO = \frac{90}{2}$

           $= 45$

$\angle BAO = \frac{90}{2}$

           $= 45$

The sum of all angles of a triangle gives 180°. So, in triangle AOB,

$\angle A + \angle O + \angle B = 180$

$45 + \angle O + 45 = 180$

$\angle O = 90$

Similarly, the other angles are the intersection are also right angles.

Therefore, it is proved that the diagonals of a square are equal and bisect each other at right angles.