# In parallelogram ABCD, the diagonals intersect at point O, ∠ = BOC140, ∠ = CAD10 and BD = 2AB.

In parallelogram ABCD, the diagonals intersect at point O, ∠ = BOC140, ∠ = CAD10 and BD = 2AB. Find the angles of the parallelogram.

Since in a parallelogram the diagonals are divided in half, then BO = DO = BD / 2, then BD = 2 * BO = 2 * DO. Since, by condition, BD = 2 * AB, then AB = BO, and therefore the ABO triangle is isosceles. Then the angle BAO = BOA. The angle BOA and BOC are adjacent angles, the sum of which is 180, then the angle BOA = BAO = 180 – 140 = 40.

Angle BAD = BAO + CAD = 40 + 10 = 50.

The sum of adjacent angles is 180, then the angle ABC = 180 – 50 = 130.

In a parallelogram, the opposite angles are equal, then the angle DC = ABC = 130, the angle BCD = BAD = 50.

Answer The angles of the parallelogram are 50 and 130.

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