Question

if the diagonals of a quadrilateral are equal and bisect each other at right angle then prove that it is a square

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Answer 1

Answer:

Proof given below:

Step-by-step explanation:

1. Let us consider that the diagonal length is x

2. If they bisect each other, the length of that will be x/2

3. Using pythagoras theorem

    => s1 (as in side 1) = $\sqrt{(x/2)^{2} + (x/2)^{2} }$

    => s1 = x/2 $\sqrt{2}$

4. Similarly,

    => s2 = x/2 $\sqrt{2}$

    => s3 = x/2 $\sqrt{2}$

    => s4 = x/2 $\sqrt{2}$

5. As s1=s2=s3=s4 and diagonals perpendicularly bisect each other, the given shape is a square. Hence Proved.

Answer 2

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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= 900    ..........(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= 1800

∠B+∠B= 1800

∠B= 900

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