In the rectangle ABCD, it is known that the angle BCA: angle DCA = 1: 5, AC = 18 cm.

In the rectangle ABCD, it is known that the angle BCA: angle DCA = 1: 5, AC = 18 cm. Find the distance from point C to diagonal BD.

Let the value of the angle BCA = X0, then, by condition, the angle DCA = 5 * X0.

Angle ВСD = 90, then X + 5 * X = 90.

6 * X = 90.

X = 90/6 = 15.

Angle DCA = 5 * 15 = 75.

Since the diagonals of the rectangle are equal and last in half at the point of their intersection, then OS = OS = AC / 2 = 18/2 = 9 cm, and therefore the triangle COD is isosceles, then the angle CDO = DCO = 75.

Angle COD = 180 – 75 – 75 = 30. Then in a right-angled triangle SON, the leg CH lies opposite angle 30, which means CH = OC / 2 = 9/2 = 4.5 cm.

Answer: The distance from point C to diagonal BD is 4.5 cm.