Question

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it
is a rhombus.
Show that the diagonals of a square are equal and bisect each other at right angles.​

Answer 1

Answer:

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  

 

∠B+∠B= 180  

 

∠B= 90  

 

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

 

So, ABCD is a square.

Hence proved.

Answer 2

Step-by-step explanation:

Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.

In △AOB and △AOD

DO=OB ∣ O is the midpoint

AO=AO ∣ Common side

∠AOB=∠AOD ∣ Right angle

So, △AOB≅△AOD

So, AB=AD

Similarly, AB=BC=CD=AD can be proved which means that ABCD is a rhombus.