Question

show that if the diagonal of a quadrilateral are equal and bisect each other at a right angle then it is a square​

Answer 1

given:

let ABCD be the quadrilateral.

diagonals are equal , AC=BD,

BISECT EACH OTHER , OA=OC ,AND OB=OD.

at right angle , ∠AOB=∠BOC=∠COD=∠AOD= 90 degree

PROOF:

prove first two triangles are similar and get the values.

substitute in the diagonals and bisect sides and you will get ab=bc=cd=ad.

hence , abcd is a square , plus the diagonals bisect rah other at 90 degree.

Answer 2

Step-by-step explanation:

Given,

Diagonals are equal

AC=BD .......(1)

and the diagonals bisect each other at right angles

OA=OC;OB=OD ...... (2)

∠AOB= ∠BOC= ∠COD= ∠AOD= 90

0

..........(3)

Proof:

Consider △AOB and △COB

OA=OC ....[from (2)]

∠AOB= ∠COB

OB is the common side

Therefore,

△AOB≅ △COB

From SAS criteria, AB=CB

Similarly, we prove

△AOB≅ △DOA, so AB=AD

△BOC≅ △COD, so CB=DC

So, AB=AD=CB=DC ....(4)

So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is parallelogram

In △ABC and △DCB

AC=BD ...(from (1))

AB=DC ...(from $$(4)$$)

BC is the common side

△ABC≅ △DCB

So, from SSS criteria, ∠ABC= ∠DCB

Now,

AB∥CD,BC is the tansversal

∠B+∠C= 180

0

∠B+∠B= 180

0

∠B= 90

0

Hence, ABCD is a parallelogram with all sides equal and one angle is 90

0

So, ABCD is a square.

Hence proved.

Hope it helps