From the middle D of side BC of an equilateral triangle ABC, a perpendicular DM to line AC is drawn

From the middle D of side BC of an equilateral triangle ABC, a perpendicular DM to line AC is drawn. Find AM if AB = 12cm.

Since triangle ABC is equilateral, all its angles are 180/3 = 60 degrees. Because D is the middle of BC, then BD = CD = 12/2 = 6 cm. Consider the triangle DMC. Because DM is perpendicular to AC, then CM is the leg adjacent to the corner C, CD is the hypotenuse. The cosine of the angle C is the ratio of CM to CD, CM can be found as the product of the hypotenuse CD and the cosine of the angle C: CM = CD * cos60 = 6 * 0.5 = 3 cm. AM = AC-CM = 12-3 = 9 cm.