Through line A of rhombus ABCD, line AP is drawn perpendicular to its plane.

Through line A of rhombus ABCD, line AP is drawn perpendicular to its plane. Find the distance from point P to lines BC, CD and BD if PA = AB = a and angle ABC = 120 °.

Since the diagonals of the rhombus divide the angles at the vertices in half, the angle ABO = ABC / 2 = 120/2 = 60.

Then, in a right-angled triangle ABO SinABO = AO / AB.

AO = AB * Sin60 = a * √3 / 2 cm.

From the right-angled triangle AРC, we determine the length of the hypotenuse OP. OР ^ 2 = AP ^ 2 + AO ^ 2 = a ^ 2 + (a * √3 / 2) ^ 2 = a ^ 2 + 3 * a ^ 2/4 = 7 * a ^ 2/4.

OР = a * √7 / 2 cm.

In the right-angled triangle ABН, the angle ABН = 180 – 120 = 60. Then, in the right-angled triangle ABН, SinAВН = AН / AB.

From the right-angled triangle of the AВР, we determine the length of the hypotenuse of the РН. PH ^ 2 = AP ^ 2 + AH ^ 2 = a ^ 2 + (a * √3 / 2) ^ 2 = a ^ 2 + 3 * a ^ 2/4 = 7 * a ^ 2/4 cm.

PH = √7 * a / 2 cm.

Answer: The distance from point P to straight lines BC, CD and BD is equal to √7 * a / 2 cm.