# Converse of Theorem 4

"If the diagonals of a quadrilateral bisect each other, then it is a parallelogram."

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**Given:**A quadrilateral ABCD in which AO = OCÂ and BO = OD.

**To prove: **Quadrilateral ABCD is a Parallelogram.**Â **

Â

**Proof** **:**

In Î”AOB and Î”COD,

AO = OC...... (given)

DO = OB..... (given)

andâˆ AOB = âˆ DOC...... (Vertically opposite angles)

.^{.}. Î”ABO â‰… Î”COD...... (SAS Criterian)

.^{.}. âˆ ABO = âˆ ODC

.^{.}. âˆ ABD = âˆ BDC

Since these are the alternate interior angles made by the transversal BD intersecting AB and DC.

AB Ã¯Ã¯ DC

Similarly AD Ã¯Ã¯ BC

Quadrilateral ABCD is a parallelogram.

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**Given: ** Ã¯Ã¯^{gm } ABCD, in whichÂ DM = NB

**To Prove:** ANCM is a parallelogram.

**Construction:** Join AC.

Â

**Proof** **:**

As the diagonals of the Ã¯Ã¯^{gm} bisect each other

AO = OC and DO = OB

And DM = BN

bisect each other

AO = OC and DO = OB

And DM = BN

Diagonals bisect each other

AO = OC and DO = OB

And DM = BN

.^{.}. DO - DM = OB - BN

.^{.}. MO = ON

Now in quadrilateral ANCM, we have

AO = OC and MO = ON

.^{.}. diagonals of this quadrilateral bisect each other.

.^{.}. ANCM is a parallelogram.