Question

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

Answer 1

Given :-

A square ABCD

AC and BD are its diagonals

To prove :-

(I) AC = BD

(II) AC and BD bisects at 90

Proff :-

(I)

In triangles ABC and BAD

AB = BA ( Common )

Angle B = Angle A ( Each 90 as all angles of square are 90 )

BC = AD ( Sides of square are equal )

Therefore by SAS Congrency

Triangle ABC is congrent to Triangle BAD

AC = BD ( C.P.C.T)

(II)

ABCD is square

ABCD is also a parallelogram ( Square is always a parallelogram )

Therefore

Its diagonals bisects each other

So ,

OA = OC

OB = OD

Now ,

In traingles AOB and COB

OA = OC ( Diagonals biscuits each other )

AB = BC ( Sides of square )

OB = OB ( Common )

Therefore by SAS Congrency

Triangle AOB is congrent to Triangle COB

Angle 1 = Angle 2 ( By C.P.C.T)

Now,

Angle 1 + Angle 2 = 180 ( Linear pair)

Angle 1 + Angle 1 = 180 ( As Angle 1 = Angle 2 )

2 Angle 1 = 180

$angle \: 1 = \frac{180}{2}$

Angle 1 = 90

So

Angle 1 = Angle 2 = 90

Angle 1 = Angle 3 = 90 ( Vertically opposite Angles )

Angle 2 = Angle 4 = 90 ( Vertically opposite Angles )

Therefore diagonals bisects at 90

Hence proffed

Note :-

Figure in attachment