Quadrangle ABCD-rhombus. The bisector BK of the angle ABD divides the side AD in half.

Quadrangle ABCD-rhombus. The bisector BK of the angle ABD divides the side AD in half. Find: 1) the corners of the rhombus; 2) the smaller diagonal of the rhombus if its side is 10 cm.

Since the bisector of the angle ABD we divide the side AD in half, then it is both the median and the height of the triangle ABD, and therefore AB = BD = AD = 10 cm, and the triangle ABD is equilateral

In an equilateral triangle, the internal angles are 60, then the angle BAD = ABD = ADB = 60.

The sum of the adjacent angles of the rhombus is 180, then the angle ABC = ADC = 180 – 60 = 120.

Answer: The angles of the rhombus are 60, 120, 60, 120, the smaller diagonal is 10 cm.