# Theorem 4 â€“ In a Parallelogram, the Diagonals Bisect Each Other

**Given: ** In a parallelogram ABCD, the diagonalsÂ AC andÂ BD intersect each other at O

**To prove: **AO = OC and BO = OD

Â

**Proof: **In Î”AOB and Î”COD, we have

Â âˆ BAO = âˆ OCD......(Alternate interior angles ofÂ AB Ã¯Ã¯ DC)

Â âˆ ABO = âˆ CDO...... (alternate interior angles)

Â AB = DCÂ ...... (opposite sides of Ã¯Ã¯ sides)

.^{.}. Î”AOB â‰… Î”COD (ASA Congruence Criteria)

Â .^{.}.Â AO = OC and DO = OB..... (CPCT)

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The diagonals of a parallelogram ABCD intersect at O. A line through O intersects AB at X and DC at Y. Show that OX = OY.

ABCD is a parallelogram

.^{.}. AB Ã¯Ã¯ DC

Also BD is a transversal of AB Ã¯Ã¯ DC

.^{.}. âˆ 1 = âˆ 2[ Alternate Interior angles are equal]

Now in Î”BXO and Î”DYO, we have

âˆ 1 = âˆ 2...... (Alternate Interior angles)

âˆ 3 = âˆ 4...... (Vertically opposite angles)

BO = OD ...... (diagonals bisects each other)

.^{.}.Â Â Î”BXO â‰… Î”DYOÂ Â (ASA Congruence Condition)

OX = OYÂ (CPCT)

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