Given,

Diagonals are equal

Diagonals are equalAC=BD .......(1)

Diagonals are equalAC=BD .......(1)and the diagonals bisect each other at right angles

Diagonals are equalAC=BD .......(1)and the diagonals bisect each other at right anglesOA=OC;OB=OD ...... (2)

Diagonals are equalAC=BD .......(1)and the diagonals bisect each other at right anglesOA=OC;OB=OD ...... (2)∠AOB= ∠BOC= ∠COD= ∠AOD= 90

Diagonals are equalAC=BD .......(1)and the diagonals bisect each other at right anglesOA=OC;OB=OD ...... (2)∠AOB= ∠BOC= ∠COD= ∠AOD= 90 0

Diagonals are equalAC=BD .......(1)and the diagonals bisect each other at right anglesOA=OC;OB=OD ...... (2)∠AOB= ∠BOC= ∠COD= ∠AOD= 90 0 ..........(3)

Proof:

Proof:Consider △AOB and △COB

Proof:Consider △AOB and △COBOA=OC ....[from (2)]

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COB

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common side

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,△AOB≅ △COB

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,△AOB≅ △COBFrom SAS criteria, AB=CB

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,△AOB≅ △COBFrom SAS criteria, AB=CBSimilarly, we prove

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,△AOB≅ △COBFrom SAS criteria, AB=CBSimilarly, we prove△AOB≅ △DOA, so AB=AD

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,△AOB≅ △COBFrom SAS criteria, AB=CBSimilarly, we prove△AOB≅ △DOA, so AB=AD△BOC≅ △COD, so CB=DC

Proof:Consider △AOB and △COBOA=OC ....[from (2)]∠AOB= ∠COBOB is the common sideTherefore,△AOB≅ △COBFrom SAS criteria, AB=CBSimilarly, we prove△AOB≅ △DOA, so AB=AD△BOC≅ △COD, so CB=DCSo, AB=AD=CB=DC ....(4)

....(4)So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is PARALLELOGRAM.

In △ABC and △DCB

In △ABC and △DCBAC=BD ...(from (1))

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCB

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCB

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0 ∠B+∠B= 180

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0 ∠B+∠B= 180 0

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0 ∠B+∠B= 180 0

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0 ∠B+∠B= 180 0 ∠B= 90

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0 ∠B+∠B= 180 0 ∠B= 90 0

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,AB∥CD,BC is the tansversal∠B+∠C= 180 0 ∠B+∠B= 180 0 ∠B= 90 0

In △ABC and △DCBAC=BD ...(from (1))AB=DC ...(from $$(4)$$)BC is the common side△ABC≅ △DCBSo, from SSS criteria, ∠ABC= ∠DCBNow,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.

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