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**Hint:**Let us suppose the quadrilateral ABCD with diagonal bisect at point O . So first let us prove that the triangle AOD and triangle COB is congruent and triangle DOC and triangle BOA is congruent by the SAS congruence rule . So from this Angle OAD and Angle OCB are alternate angles and are equal and Angle ABO and Angle CDO are alternate angles and are equal using this we can prove this

**Complete step-by-step answer:**As it is given that the ABCD is an quadrilateral with it diagonal AC and BD which intersect at the point O , and It is also given that the diagonal bisect each other , that is

OA=OC and OB=OD

In triangle AOD and COB

OA=OC It is given in the question

Angle AOD = Angle COB because it is vertical opposite

OD=OB It is given in the question

Hence the triangle AOD and triangle COB is congruent by the SAS congruence rule .

So from this we can say that angle OAD = angle OCB

Similarly, we for triangle AOB and the triangle COD

OD=OB It is given in the question

Angle DOC = Angle BOA because it is vertical opposite

OA = OC It is given in the question

Hence the triangle DOC and triangle BOA is congruent by the SAS congruence rule .

So from this we can say that angle ODC = angle OBA

For lines AB and CD and the transversal line BD,

Angle ABO and Angle CDO are alternate angles and are equal.

hence both Lines are parallel ${\text{AB}}\left\| {{\text{CD}}} \right.$

For lines AD and BC, and the transversal line AC,

Angle OAD and Angle OCB are alternate angles and are equal.

hence both Lines are parallel ${\text{AD}}\left\| {{\text{BC}}} \right.$

Thus, in quadrilateral ABCD, both pairs of opposite sides are parallel.

**Hence ABCD is a parallelogram.**

**Note:**There are five ways to find out that two triangles are congruent are SSS, SAS, ASA, AAS and HL ( hypotenuse ,leg

ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle then O is equidistant from A, B, and C

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