Prove that the point of intersection of the two diagonals of a parallelogram is the midpoint of both the diagonals.

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Solution

Given: A parallelogram ABCD with diagonals AC and BD intersecting at point O.

To Prove: AO = OC and BO = OD

Proof:

Since, ABCD is a parallelogram, AB || DC and BC || AD.

In ΔOAD and ΔOCB:

∠DAO = ∠OCB (Alternate interior angles)

∠ADO = ∠OBC (Alternate interior angles formed by parallel lines equal in magnitude)

AD = BC (Opposite sides of a parallelogram are equal in length)

As one side and two angles of a triangle are equal to the one side and two angles of the other triangle, ΔOAD ≅ ΔOCB

⇒ DO = OB and AO = OC (Corresponding parts of congruent triangles are equal)

∴O is the midpoint of DB and AC.

Thus, the point of intersection of the diagonals of a parallelogram is the midpoint of both the diagonals.

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Similar questions

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