In parallelogram ABCD, the diagonals meet at point O. a) Prove that triangle AOB is equal to triangle COD

In parallelogram ABCD, the diagonals meet at point O. a) Prove that triangle AOB is equal to triangle COD. b) It is known that AC = 10 cm, BD = 6 cm. Determine the perimeter of the triangle AOB.

In the parallelogram, the diagonals, at the point of their intersection, are divided in half, then BO = OD = BD / 2, AO = CO = AC / 2.

The opposite sides of the parallelogram are equal, AB = CD, BC = AD.

Then the triangle ABO is equal to the triangle COD on three sides, as required.

If АС = 10 cm, and ВD = 6 cm, then AO = АС / 2 = 10/2 = 5 cm, VO = ВД / 2 = 6/2 = 3 cm.

Then the perimeter of the triangle AOB will be equal to: Raov = AB + BO + AO = AB + 8 cm.

Answer: The perimeter of the ABO triangle is AB + 8 cm.