In rhombus ABCD, the height BK is drawn from the vertex of the obtuse angle B to the side AD, and the height BP

In rhombus ABCD, the height BK is drawn from the vertex of the obtuse angle B to the side AD, and the height BP is drawn to the side CD. Prove the equality of the angles KBP and BAD?

Let the value of the angle BAD = X0.

In a rhombus, the sum of adjacent angles is 180, then the value of the angle ADC = (180 – X) 0.

Consider a quadrangle КВРD.

The angle BKD and the angle BPD are straight lines, since BK and BP, by condition, are the heights of the rhombus.

The sum of the inner angles of the quadrilateral is 360, then BKD + BPD + KBP + KDP = 360.

KBP + 90 + 90 + 180 – X = 360.

Angle КBР = X + 360 – 90 – 90 – 180 = X0.

Angle КBР = BAD, which was required to be proved.