In rhombus ABCD, the bisector of angle BAC intersects side BC and diagonal BD, respectively

| March 5, 2021 | education

In rhombus ABCD, the bisector of angle BAC intersects side BC and diagonal BD, respectively, at points M and N. Find angle ANB if AMC is 120 degrees.

The diagonals of the rhombus divide the angles at the vertices in half, then the angle BAC = BCA.

Since, by condition, AM is the bisector of the angle BAC, then the angle BAM = MAC = BCA / 2.

Let the angle MAC = X0, then the angle BCA = 2 * X.

Consider a triangle AMC, the sum of the angles of which is: X + 2 * X + 120 = 180.

3 * X = 60.

X = 20.

Then the angle BAC = 2 * 20 = 400, and the angle BCD = BAD = 2 * 40 = 80.

The opposite angles of the rhombus are equal, and the sum of all the angles is 360.

Then the angle ABC = (360 – 80 – 80) / 2 = 100.

Angle AED = ABN = ABC / 2 = 100/2 = 50.

Then the angle ABN = 180 – BAN – ABN = 180 – 20 – 50 = 110.

Answer: Angle ABN = 110.



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