In parallelogram ABCD, the diagonals intersect at point O. angle BOC = 110 degrees, angle CAD
| August 3, 2021 | education
In parallelogram ABCD, the diagonals intersect at point O. angle BOC = 110 degrees, angle CAD = 40 degrees and BD = 2AB. find the angles of the parallelogram.
The diagonals of the parallelogram, at the point of their intersection, are divided in half, then BO = DO = BD / 2, and since by condition, AB = 2 * BD, then AB = BO, and then the ABO triangle is isosceles with the base AO, and then the angle BAO = BOA.
The BOA angle is adjacent to the BOC angle, the sum of which is 180, then the BOA angle = BAO = (180 – 110) = 70.
Angle BAC = BAO + OAD = 70 + 40 = 110.
The sum of the adjacent angles of the parallelogram is 180, then the angle ABC = 180 – 110 = 70.
Answer: 110, 70.
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