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

In rhombus ABCD, the bisector of angle BAC intersects side BC and diagonal BD, respectively, at points M and N. Find angle BNM if angle ACM = 12 °

In rhombus ABCD, diagonals AC and BD meet at point O at an angle of 90 °. From rectangular ∆BOC:

∠CBO = 90 ° – ∠OCB = 90 ° – ∠ACM = 90 ° – 12 ° = 78 °;

The sides of the rhombus ABCD are equal to each other, ∆ABC is isosceles, and:

∠CAB = ∠ACB = ∠ACM = 12 °;

∠ABC = 180 ° – ∠CAB – ∠ACB = 180 ° – 12 ° – 12 ° = 156 °;

By the condition of the problem:

∠MAB = ½ * ∠CAB = ½ * 12 ° = 6 °;

From ∆AMB:

∠AMB = 180 ° – ∠MAB – ∠ABC = 180 ° – 6 ° – 156 ° = 18 °;

From ∆BNM:

∠BNM = 180 ° – ∠NMB – ∠MBN = 180 ° – ∠AMB – ∠CBO = 180 ° – 18 ° – 78 ° = 84 °;

Answer: ∠BNM = 84 °